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Refraction Through A Lens: ICSE Class 10 Physics notes
Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 5 Refraction Through A Lens: ICSE Class 10 Physics (Concise/Selina Workbook) which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed.
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Summary
A lens is a clear material, often glass, with curved surfaces that bend light passing through it. There are two main types of lenses. A convex lens is thicker in its middle and thinner at its edges; it causes light rays to come together, so it is also called a converging lens. A concave lens is thinner in its middle and thicker at its edges; it causes light rays to spread apart, so it is called a diverging lens. Lenses can have different shapes, like bi-convex (both sides curved outwards) or plano-concave (one side flat, one curved inwards).
The way a lens bends light can be understood by thinking of it as a collection of tiny prisms. A convex lens acts like prisms with their bases towards the center, bending light inwards. A concave lens acts like prisms with their bases towards the edges, bending light outwards. Several terms describe lens properties. The principal axis is an imaginary straight line passing through the center of the lens. The optical centre is a point on this axis within the lens where light passes through without bending. Lenses have two principal foci. For a convex lens, parallel light rays meet at the second focal point after passing through. For a concave lens, parallel rays appear to spread from the second focal point. The distance from the optical centre to this focal point is the focal length.
To find where an image will form, we can draw ray diagrams using a few key rays. A ray passing through the optical centre goes straight. A ray parallel to the principal axis passes through (or appears to come from) the second focal point after refraction. A ray passing through the first focal point (or directed towards it) becomes parallel to the principal axis after refraction. Images can be real, meaning they can be formed on a screen, or virtual, meaning they cannot. Real images are usually inverted (upside down), while virtual images are erect (upright).
A convex lens can form different types of images depending on the object’s position. If the object is very far, the image is real, inverted, and tiny, formed at the focus. As the object moves closer, the real, inverted image moves further away and gets larger. If the object is between the focus and the lens, the image becomes virtual, erect, and magnified, appearing on the same side as the object. This is how a magnifying glass works. A concave lens always forms a virtual, erect, and diminished (smaller) image, located between the object and the lens on the same side.
A sign convention helps in calculations using the lens formula, which relates object distance (u), image distance (v), and focal length (f). Magnification (m) tells how large or small the image is compared to the object. The power of a lens measures its ability to bend light and is the reciprocal of its focal length in meters; its unit is the dioptre. Convex lenses have positive power, and concave lenses have negative power. Lenses are used in spectacles to correct vision, in cameras to form images on sensors, in telescopes to view distant objects, and in microscopes to see tiny things.
Workbook solutions (Concise/Selina)
Exercise A
MCQ
1. The given figure shows a kind of a lens. It is a :
(a) convex lens
(b) biconvex lens
(c) concave lens
(d) concavo-convex
Answer: (d) concavo-convex
2. A plano-concave lens has :
(a) one surface convex and the other surface plane.
(b) one surface concave and the other surface convex
(c) one surface plane and the other surface concave
(d) none of the above
Answer: (c) one surface plane and the other surface concave
3. The incorrect statement is :
(a) both convexo-concave and concavo-convex lenses have one surface concave and the other surface convex.
(b) a convexo-concave lens is thicker in the middle and has a converging action.
(c) A concavo-convex lens is thicker in the middle and has a converging action.
(d) a convexo-concave lens is thinner in the middle and has a diverging action.
Answer: (b) a convexo-concave lens is thicker in the middle and has a converging action.
4. A convex lens in its upper part has a prism with its base ________ and a concave lens in its upper part has a prism with its base __________
(a) downward, upward
(b) upward, downward
(c) upward, upward
(d) downward, downward
Answer: (a) downward, upward
5. The correct differences between a convex and concave lens are :
(1) convex lens is thin in the middle and concave lens is thick in the middle.
(2) convex lens converges the incident rays towards the principal axis whereas concave lens diverges
(3) convex lens has a virtual focus and concave lens has a real focus.
(a) (1)
(b) (3)
(c) (1), (2) and (3)
(d) (2)
Answer: (d) (2)
6. A ray of light after refraction through a lens emerges parallel to the principal axis of the lens. The incident ray passes through :
(a) its optical centre
(b) its first focus
(c) its second focus
(d) the centre of curvature of the first surface.
Answer: (b) its first focus
7. A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from:
(a) its first focus
(b) its optical centre
(c) its second focus
(d) the centre of curvature of its second surface.
Answer: (c) its second focus
8. A lens is called equi-convex or equi-concave when :
(a) the radius of curvature of first surface of lens is greater than the radius of curvature of the second surface.
(b) the radius of curvature of the second surface of the lens is greater than the radius of curvature of the first surface.
(c) the radius of curvature of the two surfaces of the lens are equal.
(d) none of the above
Answer: (c) the radius of curvature of the two surfaces of the lens are equal.
9. The focal length of a lens depends upon :
(a) the refractive index of the material of lens relative to its surrounding medium
(b) the amount of light entering the lens
(c) the radius of curvature of the two surfaces of lens
(d) both (a) and (c)
Answer: (d) both (a) and (c)
10. If a lens is placed in water instead of air, its focal length:
(a) increases
(b) decreases
(c) remains the same
(d) can both increase or decrease
Answer: (a) increases
11. If half part of a convex lens is covered, the focal length ________ but the intensity of the image ________
(a) increases, decreases
(b) decreases, increases
(c) does not change, increases
(d) does not change, decreases
Answer: (d) does not change, decreases
12. The focal length of a thin convex lens is ________ than that of a thick lens.
(a) less
(b) more
(c) same
(d) none of the above
Answer: (b) more
13. The radius of the sphere whose part is the lens’ surface is called the ________ of that surface.
(a) centre of curvature
(b) focal length
(c) pole
(d) radius of curvature
Answer: (d) radius of curvature
14. Identify the incorrect statement :
(a) if the medium on both sides of a lens is same, its first and second focal lengths are equal, i.e. f₁ = f₂ (numerically).
(b) when we say focal length of a lens, we mean the first focal length of the lens.
(c) a convex lens has a real focus.
(d) a concave lens has a virtual focus.
Answer: (b) when we say focal length of a lens, we mean the first focal length of the lens.
Very Short Answer Type Questions
1. What is a lens?
Answer: A lens is a transparent refracting medium bounded by either two spherical surfaces, or one spherical surface and the other surface plane.
2. Which lens is converging: (i) an equiconcave lens or an equiconvex lens. (ii) a concavo-convex lens or a convexo-concave lens?
Answer: (i) An equiconvex lens is converging. (ii) A concavo-convex lens is converging.
3. State the condition when a lens is called equiconvex or equi-concave.
Answer: A lens is called equiconvex or equi-concave when the radii of curvature of both its surfaces are equal.
4. Define the term focal length of a lens.
Answer: The focal length of a lens is the distance of the focus (or focal point) from the optical centre of the lens.
5. What do you mean by focal plane of a lens?
Answer: A focal plane of a lens is a plane normal to the principal axis, passing through the focus.
6. State the condition for each of the following :
(a) a lens has both its focal lengths equal.
Answer: A lens has both its focal lengths equal if the medium is the same on both sides of the lens.
(b) a ray passes undeviated through the lens.
Answer: A ray passes undeviated through the lens if it is directed towards the optical centre of a thin lens.
7. Complete the following sentences:
(a) If half part of a convex lens is covered, the focal length …………….. change, but the intensity of the image ……………..
Answer: If half part of a convex lens is covered, the focal length does not change, but the intensity of the image decreases.
(b) A convex lens is placed in water. Its focal length will ……………..
Answer: A convex lens is placed in water. Its focal length will increase.
(c) The focal length of a thin convex lens is …………….. than that of a thick convex lens.
Answer: The focal length of a thin convex lens is more than that of a thick convex lens.
Short Answer Type Questions
1. Name the two kinds of lens. Draw diagrams to illustrate them.
Answer: Lenses are of two kinds:
(1) convex or converging lens, and
(2) concave or diverging lens.
2. State difference between a convex and a concave lens in their (a) appearence, and (b) action on the incident light.
Answer: (a) In appearance:
A convex lens is thick in its middle and thin at the periphery.
A concave lens is thick at its periphery and thin in the middle.
(b) In action on incident light:
A convex lens converges the incident rays towards the principal axis; it has a converging action on the incident light rays.
A concave lens diverges the incident rays away from the principal axis; it has a diverging action on the incident light rays.
3. Define the term principal axis of a lens.
Answer: The principal axis of a lens is the line joining the centres of curvature of the two surfaces of the lens. It can extend indefinitely on either side of the lens.
4. Explain optical centre of a lens with the help of proper diagram(s).
Answer: The optical centre is a point on the principal axis of the lens. For a thick lens, a ray of light directed towards this point emerges parallel to its direction of incidence, though slightly displaced. For a thin lens, the lateral shift is small enough to be ignored, so a ray of light directed towards the optical centre of a thin lens can be considered to pass through the lens undeviated and undisplaced. The optical centre of a thin lens is the point on the principal axis of a lens such that a ray of light directed towards it, passes undeviated through it.
5. A ray of light incident at a point on the principal axis of a convex lens passes undeviated through the lens. (a) What special name is given to this point on the principal axis ? (b) Draw a labelled diagram to support your answer in part (a).
Answer: (a) The special name given to this point on the principal axis is the optical centre.
(b)
6. Define the term principal foci of a convex lens and illustrate your answer with the aid of proper diagrams.
Answer: A lens has two principal foci, one on either side of the lens.
For a convex lens, the first focal point (F₁) is a point on the principal axis such that the rays of light coming from it, after refraction through the lens, become parallel to the principal axis of the lens.
The second focal point (F₂) for a convex lens is a point on the principal axis such that the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it.
7. Define the term principal foci of a concave lens and show them with the help of proper diagrams.
Answer: A lens has two principal foci, one on either side of the lens.
For a concave lens, the first focal point (F₁) is a point on the principal axis such that the incident rays of light appearing to meet at it, after refraction from the lens become parallel to the principal axis of the lens.
The second focal point (F₂) for a concave lens is a point on the principal axis such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.
8. Draw a diagram to represent the second focus of a concave lens.
Answer:
9. Draw a diagram to represent the second focus of a convex lens.
Answer:
Long Answer Type Questions
1. Out of the two lenses, one concave and the other convex, state which one will show divergent action on a light beam. Draw diagrams to illustrate your answer.
Answer: A concave lens will show divergent action on a light beam. A concave lens diverges (or spreads) the parallel rays as if they are coming from a common point F situated on the side of rays incident on the lens. Therefore, a concave lens has a diverging action on the incident light rays.
2. Show by a diagram the refraction of two light rays incident parallel to the principal axis on a convex lens by treating it as a combination of a glass slab and two triangular glass prisms.
Answer: A convex lens can be considered as being made up of a rectangular slab at the centre and one prism on either side of it. The prism in the upper part of the convex lens bends an incident ray parallel to the principal axis downwards towards its base, while the prism in the lower part bends another incident ray parallel to the principal axis upwards towards its base. The central part, which is a parallel sided glass slab, passes a ray normally incident on it, undeviated. Thus, the set of prisms forming a convex lens converges the parallel rays.
3. Show by a diagram, the refraction of two light rays incident parallel to the principal axis on a concave lens by treating it as a combination of a glass slab and two triangular glass prisms.
Answer: A concave lens can be considered as being made up of a rectangular slab at the centre and one prism on either side of it. The prism in the upper part of the concave lens bends an incident ray parallel to the principal axis upwards towards its base, while the prism in the lower part bends another incident ray parallel to the principal axis downwards towards its base. The central part, which is a parallel sided glass slab, passes a ray normally incident on it, undeviated. Thus, the set of prisms forming a concave lens diverges the parallel rays.
4. How does the action of a convex lens differ from that of a concave lens on a parallel beam of light incident on them ? Draw diagrams to illustrate your answer.
Answer: A convex lens has a converging action on a parallel beam of incident light rays, bringing the rays closer to a point F. A concave lens has a diverging action on a parallel beam of incident light rays, spreading the parallel rays as if they are coming from a common point F.
5. A ray of light, after refraction through a concave lens emerges parallel to the principal axis. (a) Draw a ray diagram to show the incident ray and its corresponding emergent ray. (b) The incident ray when produced meets the principal axis at a point F. Name the point F.
Answer: (a) The ray diagram shows an incident ray directed towards the first focal point of the concave lens, which after refraction emerges parallel to the principal axis.
(b) The point F where the incident ray, when produced, meets the principal axis is the first focal point (F₁) of the concave lens.
6. A ray of light after refraction through a convex lens emerges parallel to the principal axis. (a) Draw a ray diagram to show it. (b) The incident ray passes through a point F on the principal axis. Name the point F.
Answer: (a) The ray diagram shows an incident ray passing through the first focal point of the convex lens, which after refraction emerges parallel to the principal axis.
(b) The point F on the principal axis through which the incident ray passes is the first focal point (F₁) of the convex lens.
7. A beam of light incident on a convex lens parallel to its principal axis converges at a point F on the principal axis. Name the point F. Draw a ray diagram to show it.
Answer: The point F on the principal axis where the beam of light converges is the second focal point (F₂) of the convex lens.
8. A beam of light incident on a thin concave lens parallel to its principal axis diverges and appears to come from a point F on the principal axis. Name the point F. Draw a ray diagram to show it.
Answer: The point F on the principal axis from which the beam of light appears to diverge is the second focal point (F₂) of the concave lens.
9. A parallel oblique beam of light falls on a (i) convex lens, (ii) concave lens. Draw a diagram in each case to show refraction of light through the lens.
Answer: (i) When a parallel oblique beam of light falls on a convex lens, it converges at some point B in the second focal plane of the lens.
(ii) When a parallel oblique beam of light falls on a concave lens, after refraction it appears to diverge from a point B in the second focal plane of the lens.
10. The diagram below shows a lens as a combination of a glass slabs and two prisms.
(i) Name the lens formed by the combination.
Answer: The lens formed by the combination is a convex lens.
(ii) What is the line XX’ called ?
Answer: The line XX’ is called the principal axis.
(iii) Complete the ray diagram and show the path of the incident ray AB after passing through the lens.
Answer: The incident ray AB, which is parallel to the principal axis XX’, after passing through the convex lens will refract and pass through the second focal point F on the principal axis.
(iv) The final emergent ray will either meet XX’ at a point or appear to come from a point on XX’. Label the point as F. What is this point called?
Answer: The final emergent ray will meet XX’ at a point F. This point F is called the second focal point (or principal focus) of the convex lens.
11. The diagram below shows a lens as a combination of a glass slab and two prisms.
(i) Name the lens formed by the combination.
Answer: The lens formed by the combination is a concave lens.
(ii) What is the line XX’ called ?
Answer: The line XX’ is called the principal axis.
(iii) Complete the path of the incident ray AB after passing through the lens.
Answer: The incident ray AB, which is parallel to the principal axis XX’, after passing through the concave lens will refract and diverge such that it appears to come from the second focal point F on the principal axis, located on the same side as the incident ray.
(iv) The final emergent ray either meets XX’ at a point or appears to come from a point on XX’. Label it as F. What is this point called?
Answer: The final emergent ray will appear to come from a point F on XX’. This point F is called the second focal point (or principal focus) of the concave lens.
12. In figures (a) and (b), F₁ and F₂ are positions of the two foci of thin lenses. Draw the path taken by the light ray AB after it emerges from each lens.
Answer: (a) In fig. (a), the lens is convex. The incident ray AB is parallel to the principal axis. After refraction, it will pass through the second focal point F₂.
(b) In fig. (b), the lens is concave. The incident ray AB is parallel to the principal axis. After refraction, it will diverge as if it is coming from the second focal point F₂ (which is on the same side as the incident ray).
13. In fig. (a) and (b), F₁ and F₂ are the two foci of thin lenses and AB is the incident ray. Complete the diagram to show the path of the ray AB after refraction through each lens.
Answer: (a) In Fig. (a), the lens is convex. The incident ray AB is passing through the first focal point F₁. After refraction, it will emerge parallel to the principal axis.
(b) In Fig. (b), the lens is concave. The incident ray AB is directed towards the first focal point F₁ (located on the opposite side of the lens). After refraction, it will emerge parallel to the principal axis.
Exercise B
MCQ
1. A ray of light parallel to the principal axis for a concave lens after refraction:
(a) passes through the second focus
(b) passes through the first focus
(c) appears to pass through the second focus
(d) appears to come from the second focus
Answer: (d) appears to come from the second focus
2. Which of the ray diagrams are correct ?
(a)
(b)
(c)
(d)
Answer: (a)
3. Identify the incorrect statement :
(a) A real image is formed due to actual intersection of the rays refracted by the lens.
(b) A virtual image can be obtained on the screen.
(c) A virtual image is erect with respect to the object.
(d) A real image is inverted with respect to the object.
Answer: (b) A virtual image can be obtained on the screen.
4. A ray diagram for a convex lens is shown below. In this manner, a convex lens is used in :
(a) camera
(b) cinema
(c) burning glass
(d) terrestrial telescope
Answer: (d) terrestrial telescope
5. For an object placed at a distance 20 cm in front of a convex lens, the image is at a distance 20 cm behind the lens. The focal length of the convex lens is:
(a) 20 cm
(b) 10 cm
(c) 15 cm
(d) 40 cm.
Answer: (b) 10 cm
6. For the object placed between the optical centre and focus of a convex lens, the image is :
(a) real and enlarged
(b) real and diminished
(c) virtual and enlarged
(d) virtual and diminished
Answer: (c) virtual and enlarged
7. A concave lens forms the image of an object which is :
(a) virtual, inverted, and diminished
(b) virtual, upright, and diminished
(c) virtual, inverted, and enlarged
(d) virtual, upright, and enlarged.
Answer: (b) virtual, upright, and diminished
8. For a convex lens when the object is at 2F₁, then image is formed at :
(a) F₂
(b) 2F₂
(c) beyond 2F₂
(d) infinity
Answer: (b) 2F₂
9. A concave lens is used in:
(a) in slide projector
(b) as magnifying glass
(c) as burning glass
(d) in Galilean telescope
Answer: (d) in Galilean telescope
10. For a concave lens, when the object is at a distance equal to the focal length of the lens, the image is formed at:
(a) optical centre
(b) second focus
(c) first focus
(d) mid point between the optical centre and second focus of the lens
Answer: (d) mid point between the optical centre and second focus of the lens
11. Assertion (A): The focal length of a lens does not change when red light is replaced by blue light.
Reason (R): The focal length of a lens depends on the refractive index of material of lens.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (c) assertion is false but reason is true
Very Short Answer Type Questions
1. A lens forms an inverted image of an object.
(a) Name the kind of lens.
(b) State the nature of the image whether real or virtual?
Answer: (a) The kind of lens is convex.
(b) The nature of the image is real.
2. (a) Name the lens which always forms an erect and virtual image.
(b) State whether the image in part (a) is magnified or diminished.
Answer: (a) The lens which always forms an erect and virtual image is a concave lens.
(b) The image in part (a) is diminished.
3. Can a concave lens form an image of size two times that of the object? Give reason.
Answer: No. The reason is that a concave lens diverges the rays incident on it, and the image is always diminished.
4. Give two characteristics of the image formed by a concave lens.
Answer: Two characteristics of the image formed by a concave lens are that it is always virtual and upright, and it is always diminished. It is also situated on the side of the object between the focus and the lens.
5. Give two characteristics of the virtual image formed by a convex lens.
Answer: Two characteristics of the virtual image formed by a convex lens are that it is erect or upright, and it is magnified. It is also formed on the same side as the object and behind the object.
6. Complete the following sentences :
(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and……
(b) An object is placed at a distance 2f from a convex lens of focal length f. The size of the image formed is ………… that of the object.
(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and …………
Answer: (a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and diminished.
(b) An object is placed at a distance 2f from a convex lens of focal length f. The size of the image formed is equal to (or same as) that of the object.
(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and magnified.
7. State whether the following statements are ‘true’ or ‘false’ by writing T/F against them.
(a) A convex lens has a divergent action and a concave lens has a convergent action.
(b) A concave lens, if kept at a proper distance from an object, can form its real image.
(c) A ray of light incident parallel to the principal axis of a lens, passes undeviated after refraction.
(d) A ray of light incident at the optical centre of a lens, passes undeviated after refraction.
(e) A concave lens forms a magnified or diminished image depending on the distance of an object from it.
Answer:
(a) F
(b) F
(c) F
(d) T
(e) F
8. Where must a point source of light be placed in front of a convex lens so as to obtain a parallel beam of light?
Answer: A point source of light must be placed at the first focus (F₁) of a convex lens to obtain a parallel beam of light.
Short Answer Type Questions
1. What are the three principal rays that are drawn to construct a ray diagram for the image formed by a lens? Draw diagrams to support your answer.
Answer: Generally, the following three principal rays are used for the construction of ray diagrams:
(1) A ray of light incident at the optical centre O of the lens passes undeviated through the lens.
(2) A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F₂ (in a convex lens) or appears to come from the second focus F₂ (in a concave lens).
(3) A ray of light passing through the first focus F₁ (in a convex lens) or directed towards the first focus F₁ (in a concave lens), emerges parallel to the principal axis after refraction.
2. In the diagrams below, XX’ represents the principal axis, O the optical centre and F the focus of the lens. Complete the path of the rays A and B as they emerge out of the lens.
(a)
(b)
Answer:
(a)
(b)
3. Distinguish between a real and a virtual image.
Answer: The distinction between a real and a virtual image is as follows:
| Real Image | Virtual Image |
| 1. A real image is formed due to actual intersection of the rays refracted by the lens. | 1. A virtual image is formed when the rays refracted by the lens appear to meet if they are produced backwards. |
| 2. A real image can be obtained on a screen. | 2. A virtual image can not be obtained on a screen. |
| 3. A real image is inverted with respect to the object. | 3. A virtual image is erect with respect to the object. |
| Example: The image of a distant object formed by a convex lens on a screen | Example: The image of an object formed by a concave lens as seen by the eye. |
4. A lens forms an upright and magnified image of an object.
(a) Name the lens.
(b) Draw a labelled ray diagram to show the image formation.
Answer: (a) The lens is a convex lens.
(b) To show the image formation, an object AB is placed on the principal axis of a convex lens between its optical centre O and the first focus F₁. A ray AD from point A, incident parallel to the principal axis, after refraction through the lens passes through the second focus F₂. Another ray AO from point A, incident at the optical centre O of the lens, passes undeviated. These two refracted rays do not meet but appear to diverge from a point A’ when produced backwards. Thus, A’ is the virtual image of point A. A’B’ is the virtual, erect, and magnified image of the object AB, formed on the same side as the object and behind it.
5. In each of the following cases, where must an object be placed in front of a convex lens so that the image formed is
(a) at infinity,
(b) of the same size as the object,
(c) inverted and enlarged,
(d) upright and enlarged ?
Answer: For a convex lens, the object must be placed:
(a) at infinity: at the first focus (F₁).
(b) of the same size as the object: at 2F₁ (twice the focal length from the lens).
(c) inverted and enlarged: between F₁ and 2F₁.
(d) upright and enlarged: between the optical centre (O) and the first focus (F₁).
6. Complete the following table :
Answer:
| Type of Lens | Position of Object | Nature of Image | Size of Image |
| Convex | Between optical centre and focus | Virtual and upright | Magnified |
| Convex | At focus | Real and inverted | Highly magnified |
| Concave | At infinity | Virtual and upright | Highly diminished |
| Concave | At any distance (between infinity and optical centre) | Virtual and upright | Diminished |
Long Answer Type Questions
1. Study the diagram given below.
(a) Name the lens LL’.
(b) What are the points O and O’ called ?
(c) Complete the diagram to form the image of the object AB.
(d) State the three characteristics of the image.
(e) Name a device in which this action of lens is used.
Answer: (a) The lens LL’ is a concave lens.
(b) The point O is the optical centre of the lens. The point O’ (where B is located) is a point on the principal axis representing the position of the base of the object AB.
(c) To complete the diagram: A ray from A parallel to the principal axis, after refraction by the concave lens LL’, appears to diverge from the second focus F₂ (which would be on the same side as the object). Another ray from A passing through the optical centre O goes undeviated. The intersection of the backward extension of the diverged ray and the undeviated ray forms the virtual image point A’. A’B’ is the image.
(d) The three characteristics of the image are that it is virtual, erect (or upright), and diminished.
(e) This action of a concave lens is used in spectacles for short-sighted persons or as the eye lens in a Galilean telescope.
2. Study the diagram below.
(i) Name the lens LL’.
(ii) What are the points O and O’ called ?
(iii) Complete the diagram to form the image of the object AB.
(iv) State three characteristics of the image.
Answer: (i) The lens LL’ is a convex lens.
(ii) The point O is the optical centre of the lens. The point O’ (where B is located) is a point on the principal axis representing the position of the base of the object AB.
(iii) To complete the diagram: A ray from A parallel to the principal axis, after refraction by the convex lens LL’, passes through the second focus F₂ (on the opposite side). Another ray from A passing through the optical centre O goes undeviated. The intersection of these two refracted rays forms the real image point A’. A’B’ is the image.
(iv) Assuming the object AB is placed beyond 2F₁, the three characteristics of the image are that it is real, inverted, and diminished. If it were between F₁ and 2F₁, it would be real, inverted, and magnified.
3. The diagram shows an object AB and a converging lens L with foci F₁ and F₂.
(a) Draw two rays from the object AB and complete the diagram to locate the position of the image CD. Also mark on the diagram the position of eye from where the image can be viewed.
(b) State three characteristics of the image in relation to the object.
Answer:
(a)
(b) The object AB appears to be placed beyond 2F₁. In this case, the image CD will be real, inverted, and diminished (smaller than the object).
4. The diagram shows the position of an object OA in relation to a converging lens L whose foci are at F₁ and F₂.
(i) Draw two rays to locate the position of the image.
(ii) State the position of the image with reference to the lens.
(iii) Describe three characteristics of the image.
(iv) Describe how the distance of the image from the lens and its size change as the object is moved towards F₁.
Answer:
(i)
(ii) The object OA is placed between F₁ and 2F₁. The image A’O’ is formed beyond 2F₂ on the other side of the lens.
(iii) The three characteristics of the image are that it is real, inverted, and magnified.
(iv) As the object is moved from its position (between F₁ and 2F₁) towards F₁, the image moves further away from 2F₂ (towards infinity) on the other side of the lens, and its size (magnification) increases.
5. A converging lens forms the image of an object placed in front of it, beyond 2F₁ of the lens.
(a) Where is the object placed ? (This question seems to ask for image position given the object position)
(b) Draw a ray diagram to show the formation of the image.
(c) State three characteristics of the image.
Answer: (a) The object is placed beyond 2F₁ (twice the focal length) on the principal axis of the converging (convex) lens. The image is formed between F₂ and 2F₂ on the other side of the lens.
(b)
(c) The three characteristics of the image are that it is real, inverted, and diminished.
6. A convex lens forms an image of an object equal to the size of the object.
(a) Where is the object placed in front of the lens ?
(b) Draw a diagram to illustrate it.
(c) State two more characterstics of the image.
Answer: (a) The object is placed at 2F₁ (at a distance equal to twice the focal length) in front of the convex lens.
(b) To illustrate with a diagram: An object AB is placed on the principal axis of a convex lens at 2F₁. A ray AD from point A, incident parallel to the principal axis, after refraction through the lens, passes through its second focus F₂ as DA’. Another ray AO from point A, incident through the optical centre O of the lens, travels undeviated as OA’. The two refracted rays DA’ and OA’ meet at point A’. Thus, A’B’ is the image of the object AB, and it is of the same size.
(c) Two more characteristics of the image are that it is real and inverted.
7. A lens forms an erect, magnified, and virtual image of an object.
(a) Name the kind of lens.
(b) Where is the object placed in relation to the lens ?
(c) Draw a ray diagram to show the formation of image.
(d) Name the device which uses this principle.
Answer: (a) The kind of lens is a convex lens.
(b) The object is placed between the optical centre O and the first focus F₁ of the convex lens (i.e., its distance from the lens is less than the focal length).
(c)
(d) This principle is used in a reading lens, which is a magnifying glass or a simple microscope.
8. A lens forms an image between the object and the lens.
(a) Name the lens.
(b) Draw a ray diagram to show the formation of such an image.
(c) State three characteristics of the image.
Answer: (a) The lens is a concave lens.
(b)
(c) Three characteristics of the image are that it is virtual, erect (or upright), and diminished.
9. Classify as real or virtual, the image of a candle flame formed on a screen by a convex lens. Draw a ray diagram to illustrate how an image is formed.
Answer: The image of a candle flame formed on a screen by a convex lens is a real image.
To illustrate with a ray diagram: A candle flame AB is placed as an object in front of a convex lens, at a distance greater than its focal length (e.g., beyond 2F₁). A ray AD from the top of the flame A, parallel to the principal axis, after refraction through the lens, passes through the second focus F₂. Another ray AO from A, passing through the optical centre O of the lens, goes undeviated. These two refracted rays intersect at a point A’ on the other side of the lens. A’B’ is the real, inverted image formed, which can be obtained on a screen placed at this position.
10. Show by a ray diagram that a diverging lens cannot form a real image of an object placed anywhere on its principal axis.
Answer: A ray AD from point A, parallel to the principal axis, after refraction by the concave lens, diverges and appears to come from the second focus F₂ (located on the same side as the object).
A ray AO from point A, passing through the optical centre O, goes undeviated. The actual refracted rays (the diverged ray appearing from F₂ and the undeviated ray through O) spread out and do not intersect. They only appear to intersect at point A’ when the first ray is produced backwards. Thus, A’B’ is a virtual image. Since the refracted rays always diverge from each other after passing through a concave lens, they can never meet at a point to form a real image.
11. Draw a ray diagram to show how a converging lens can form a real and enlarged image of an object.
Answer: A converging lens (convex lens) forms a real and enlarged (magnified) image when the object is placed between its first focal point F₁ and twice its focal length (2F₁).
Ray diagram: An object AB is placed on the principal axis of a convex lens at a point between F₁ and 2F₁. A ray AD from point A, incident parallel to the principal axis, after refraction from the lens, passes through its second focus F₂. Another ray AO from point A, incident towards the optical centre O, passes undeviated through the lens. The two refracted rays meet at a point A’ beyond 2F₂ on the other side of the lens. A’B’ is the real, inverted, and magnified image of the object AB.
12. A lens forms an upright and diminshed image of an object placed at its focal point. Name the lens and draw a ray diagram to show the formation of the image.
Answer: The lens is a concave lens. (A concave lens always forms an upright and diminished image. If the object is placed at its focal point, the image is formed at half the focal length, between the optical centre and the focus).
Ray diagram: Let AB be an object placed at the first focal point F₁ of a concave lens (meaning its distance from the lens is equal to the focal length).
- A ray AD from point A, parallel to the principal axis, after refraction, appears to diverge from the second focus F₂ (on the same side as the object).
- A ray AO from point A, passing through the optical centre O, goes undeviated.
These two refracted rays, when the first is produced backwards, appear to meet at A’. A’B’ is the virtual, upright, and diminished image formed between the optical centre O and the second focus F₂. Specifically, it is formed at the mid-point between O and F₂.
13. Draw a ray diagram to show how a converging lens is used as a magnifying glass to observe a small object. Mark on your diagram the foci of the lens and the position of the eye.
Answer: A converging lens (convex lens) is used as a magnifying glass when a small object is placed between its optical centre O and its first focus F₁.
Ray diagram: A tiny object AB is placed on the principal axis of a convex lens, between O and F₁.
- A ray AD from point A of the object, incident parallel to the principal axis, after refraction from the lens, passes through its second focus F₂.
- Another incident ray AO from point A, passing through the optical centre O, gets refracted without deviation.
The two refracted rays do not actually meet but appear to diverge from a point A’ when they are produced backwards. Thus, A’ is the virtual image of point A. Similarly, B’ is the virtual image of B. A’B’ is the virtual, magnified, and erect image of the object AB, formed on the same side as the object and behind it. The eye is kept very close to the lens on the side from which the diverging rays appear to come, to view this magnified image.
14. Draw a ray diagram to show how a converging lens can form an image of the sun. Hence give a reason for the term ‘burning glass’ for a converging lens used in this manner.
Answer: The sun is considered an object at infinity, so the rays of light coming from it are parallel to each other.
Ray diagram: Rays of light from the sun, incident parallel to the principal axis of a convex lens, after refraction through the lens, converge and pass through the second focus F₂ of the lens. Thus, a real, inverted, and highly diminished (almost to a point) image of the sun is formed at F₂ on the other side of the lens.
Reason for ‘burning glass’: When a convex lens is used in this manner to focus the parallel rays of light from the sun, these rays are concentrated at the focal point (F₂). If a piece of paper is kept at this focal plane, the concentration of light energy produces sufficient heat to cause the paper to burn. This is why a converging lens used in this way is termed a ‘burning glass’.
15. State the changes in the position, size and nature of the image when the object is brought from infinity up to the convex lens. Illustrate your answer by drawing ray diagrams.
Answer: The changes in the position, size, and nature of the image formed by a convex lens as the object is brought from infinity up to the lens are as follows:
When the object is at infinity (u = ∞): The image is formed at the second focus F₂, is real, inverted, and highly diminished (almost to a point).
When the object is beyond 2F₁ (u > 2f): The image is formed between F₂ and 2F₂ on the other side, is real, inverted, and diminished. As the object moves from infinity towards 2F₁, the image moves away from F₂ towards 2F₂, and its size gradually increases (but remains diminished).
When the object is at 2F₁ (u = 2f): The image is formed at 2F₂ on the other side, is real, inverted, and of the same size as the object.
When the object is between F₁ and 2F₁ (f < u < 2f): The image is formed beyond 2F₂ on the other side, is real, inverted, and magnified. As the object moves from 2F₁ towards F₁, the image moves further away from 2F₂ towards infinity, and its size (magnification) increases.
When the object is at F₁ (u = f): The image is formed at infinity, is real (or considered so as rays become parallel), inverted, and highly magnified.
When the object is between the lens (optical centre O) and F₁ (u < f): The image is formed on the same side as the object, behind the object, is virtual, erect, and magnified. As the object moves from F₁ towards the lens, the image also moves from infinity towards the lens (behind the object), and its size decreases (though it remains magnified).
16. State the changes in the position, size and nature of the image when the object is brought from infinity up to the concave lens. Illustrate your answer by drawing ray diagrams.
Answer: Irrespective of the position of the object, the image formed by a thin concave (or divergent) lens is always virtual, upright, and diminished, and it is situated on the side of the object between the focus and the lens.
The changes are as follows:
- When the object is at infinity: The image is formed at the second focus F₂ (on the same side as the object), is virtual, erect, and highly diminished (almost to a point).
- When the object moves from a large distance towards the lens: The image shifts from the focus F₂ towards the optical centre O of the lens (remaining on the same side as the object). The image remains virtual, upright, and diminished, but its size gradually increases (though it always remains smaller than the object).
(iii) When the object is at a distance equal to the focal length of the lens: The image is exactly at the mid-point between the optical centre and the second focus of the lens. It is virtual, upright, and diminished.
As the object moves from infinity up to the optical centre of the concave lens, the image moves from F₂ up to the optical centre, always remaining virtual, erect, and diminished, with its size increasing from a point to nearly the size of the object (but always smaller) if the object were at O.
Exercise C
MCQ
1. The correct lens formula is :
(a) 1/u + 1/v = 1/f
(b) 1/v – 1/u = 1/f
(c) f = uv / (u-v)
(d) f = uv / (u+v)
Answer: (c) f = uv / (u-v)
2. Identify the wrong statement.
(a) For a convex lens, u is always negative.
(b) For a convex lens, f is always positive.
(c) For a convex lens, v is positive for a virtual image and negative for a real image.
(d) For a concave lens, u, v and f all are negative.
Answer: (c) For a convex lens, v is positive for a virtual image and negative for a real image.
3. For a convex lens, the value of magnification m is :
(a) positive
(b) negative
(c) both positive as well as negative
(d) none
Answer: (c) both positive as well as negative
4. If the magnification produced by a lens is – 0·5, the correct statement is :
(a) the lens is concave
(b) the image is virtual
(c) the image is magnified
(d) the image is real and diminshed formed by a convex lens
Answer: (d) the image is real and diminshed formed by a convex lens
5. The numerical value of magnification m is always ……… for a concave lens.
(a) equal to 1
(b) less than 1
(c) more than 1
(d) variable
Answer: (b) less than 1
6. If a lens deviates a ray towards its centre, its power is ……… and if it deviates the ray away from its centre, its power is ………
(a) positive, positive
(b) negative, negative
(c) negative, positive
(d) positive, negative
Answer: (d) positive, negative
7. On reducing the focal length of a lens, its power :
(a) decreases
(b) increases
(c) does not change
(d) first increases then decreases.
Answer: (b) increases
8. The lens of power + 1.0 D is :
(a) convex of focal length 1.0 cm
(b) convex of focal length 1.0 m
(c) concave of focal length 1.0 cm
(d) concave of focal length 1.0 m.
Answer: (b) convex of focal length 1.0 m
9. Linear magnification m is given by :
(a) m = vu
(b) m = 1/v – 1/u
(c) m = v/u
(d) m = u/v
Answer: (c) m = v/u
10. If a lens is placed in water instead of air, its power :
(a) increases
(b) decreases
(c) remains the same
(d) can both increase or decrease
Answer: (b) decreases
Very Short Answer Type Questions
1. The focal length of a lens is (i) positive, (ii) negative. In each case, state the kind of lens.
Answer: (i) If the focal length of a lens is positive, the lens is convex.
(ii) If the focal length of a lens is negative, the lens is concave.
2. What information about the nature of image (i) real or virtual, (ii) erect or inverted, do you get from the sign of magnification + or – ?
Answer: (i) If the sign of magnification is positive (+), the image is virtual. If the sign of magnification is negative (–), the image is real.
(ii) If the sign of magnification is positive (+), the image is erect. If the sign of magnification is negative (–), the image is inverted.
3. How is the power of a lens related to its focal length ?
Answer: The power of a lens is expressed (or measured) in terms of the reciprocal of its focal length.
4. How does the power of a lens change if its focal length is doubled ?
Answer: If the focal length of a lens is doubled, its power gets halved.
5. How is the sign (+ or -) of power of a lens related to its divergent or convergent action ?
Answer: If a lens deviates a ray towards its centre (convergent action), its power is positive (+). If it deviates the ray away from its centre (divergent action), its power is negative (–).
6. The power of a lens is negative. State whether it is convex or concave ?
Answer: If the power of a lens is negative, it is a concave lens.
7. Which lens has more power: a thick lens or a thin lens ?
Answer: A thick lens has more power.
Short Answer Type Questions
1. State the sign convention to measure distances for a lens.
Answer: The cartesian sign convention to measure the distances in a lens is as follows:
(1) The optical centre of the lens is chosen as the origin of the coordinate system.
(2) The object is considered to be placed on the principal axis to the left of the lens.
(3) All distances are measured along the principal axis from the optical centre of the lens. The distance of an object from the lens is denoted by u, the distance of image by v and the distance of second focus (i.e., focal length of lens) by f.
(4) Distances measured in the direction of the incident ray are taken positive, while distances opposite to the direction of the incident ray are taken negative.
(5) The length above the principal axis is taken positive, while the length below the principal axis is taken negative.
2. Write the lens formula explaining the meaning of the symbols used.
Answer: The lens formula is 1/v – 1/u = 1/f.
The equation relating the distance of object (u), distance of image (v) and focal length (f) of a lens is called the lens formula. It is the same for both the convex and the concave lens.
3. What do you understand by the term magnification ? Write an expression for it for a lens, explaining the meaning of the symbols used.
Answer: The ratio of the length of image I perpendicular to the principal axis, to the length of object O, is called linear magnification. It is generally denoted by the letter m.
An expression for it for a lens is m = length of image (I) / length of object (O) = v/u, where v is the distance of the image and u is the distance of the object.
4. Define the term power of a lens. In what unit is it expressed ?
Answer: The deviation of the incident light rays produced by a lens on refraction through it, is a measure of its power. A lens which produces more deviation has more power. Power of a lens is expressed (or measured) in terms of the reciprocal of its focal length. Its unit is dioptre (symbol D).
5. A lens of power +2.5D is kept in contact with another lens of power -2.5D. What will be the power of the combination of the two lenses ? How will the combination behave ?
Answer: If two thin lenses are placed in contact, the combination has a power equal to the algebraic sum of the powers of the individual lens.
So, the power of the combination will be +2.5D + (-2.5D) = 0 D.
The combination will have zero power and it will behave like a glass plate.
Numericals
1. (a) At what position a candle of length 3 cm be placed in front of a convex lens so that its image of length 6 cm be obtained on a screen placed at a distance 30 cm behind the lens?
(b) What is the focal length of the lens in part (a)?
Answer:
(a) Position of the candle
Given:
Object height, O = +3 cm
Image height, I = -6 cm (Image is on a screen, so it is real and inverted)
Image distance, v = +30 cm (Image is behind the lens)
To find:
Object distance, u = ?
Solution:
Using the magnification formula,
m = I / O = v / u
=> -6 / 3 = 30 / u
=> -2 = 30 / u
=> u = 30 / -2
=> u = -15 cm.
Thus, the candle should be placed at a distance of 15 cm in front of the lens.
(b) Focal length of the lens
Given:
Image distance, v = +30 cm
Object distance, u = -15 cm
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/30 – 1/(-15)
=> 1/f = 1/30 + 1/15
=> 1/f = (1 + 2) / 30
=> 1/f = 3 / 30
=> 1/f = 1 / 10
=> f = +10 cm.
Thus, the focal length of the convex lens is 10 cm.
2. A concave lens forms the image of an object kept at a distance 20 cm in front of it, at a distance 10 cm on the side of the object. (a) What is the nature of the image? (b) Find the focal length of the lens.
Answer:
(a) Nature of the image
A concave lens always forms a virtual, erect and diminished image on the same side as the object.
(b) Focal length of the lens
Given:
Object distance, u = -20 cm
Image distance, v = -10 cm (Image is on the same side as the object)
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/(-10) – 1/(-20)
=> 1/f = -1/10 + 1/20
=> 1/f = (-2 + 1) / 20
=> 1/f = -1 / 20
=> f = -20 cm.
Thus, the focal length of the concave lens is 20 cm.
3. The focal length of a convex lens is 25 cm. At what distance from the optical centre of the lens an object be placed to obtain a virtual image of twice the size?
Answer:
Given:
Focal length, f = +25 cm (convex lens)
Magnification, m = +2 (Image is virtual and twice the size)
To find:
Object distance, u = ?
Solution:
First, using the magnification formula,
m = v / u
=> +2 = v / u
=> v = 2u
Now, using the lens formula,
1/f = 1/v – 1/u
=> 1/25 = 1/(2u) – 1/u
=> 1/25 = (1 – 2) / (2u)
=> 1/25 = -1 / (2u)
=> 2u = -25
=> u = -12.5 cm.
Thus, the object should be placed at a distance of 12.5 cm in front of the lens.
4. Where should an object be placed in front of a convex lens of focal length 0.12 m to obtain a real image of size three times the size of the object, on the screen?
Answer:
Given:
Focal length, f = +0.12 m (convex lens)
Magnification, m = -3 (Image is real and three times the size)
To find:
Object distance, u = ?
Solution:
First, using the magnification formula,
m = v / u
=> -3 = v / u
=> v = -3u
Now, using the lens formula,
1/f = 1/v – 1/u
=> 1/0.12 = 1/(-3u) – 1/u
=> 1/0.12 = (-1 – 3) / (3u)
=> 1/0.12 = -4 / (3u)
=> 3u = -4 × 0.12
=> 3u = -0.48
=> u = -0.16 m.
Thus, the object should be placed at a distance of 0.16 m (or 16 cm) in front of the lens.
5. An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form an image of the object on the screen placed at a distance of 75 cm from the lens. Find: (i) the focal length of the lens, and (ii) the magnification.
Answer:
Given:
Distance between object and screen = 1.0 m = 100 cm.
Image distance, v = +75 cm (Image is on the screen).
The object distance u must be negative. The distance is |u| + v.
|u| + 75 cm = 100 cm
=> |u| = 25 cm
So, object distance, u = -25 cm.
(i) Focal length of the lens
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/75 – 1/(-25)
=> 1/f = 1/75 + 1/25
=> 1/f = (1 + 3) / 75
=> 1/f = 4 / 75
=> f = 75 / 4
=> f = +18.75 cm.
Thus, the focal length of the lens is 18.75 cm.
(ii) The magnification
To find:
Magnification, m = ?
Solution:
Using the magnification formula,
m = v / u
=> m = 75 / (-25)
=> m = -3.
Thus, the magnification is -3.
6. A lens forms the image of an object placed at a distance of 15 cm from it, at a distance of 60 cm in front of it. Find: (i) the focal length, (ii) the magnification, and (iii) the nature of image.
Answer:
Given:
Object distance, u = -15 cm
Image distance, v = -60 cm (Image is in front of the lens, so it’s virtual)
(i) The focal length
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/(-60) – 1/(-15)
=> 1/f = -1/60 + 1/15
=> 1/f = (-1 + 4) / 60
=> 1/f = 3 / 60
=> 1/f = 1 / 20
=> f = +20 cm.
Thus, the focal length is 20 cm. Since f is positive, the lens is a convex lens.
(ii) The magnification
To find:
Magnification, m = ?
Solution:
Using the magnification formula,
m = v / u
=> m = (-60) / (-15)
=> m = +4.
Thus, the magnification is +4.
(iii) The nature of image
Since the magnification is positive (m = +4) and greater than 1, the image is virtual, erect, and magnified.
7. A lens forms the image of an object placed at a distance of 45 cm from it on a screen placed at a distance 90 cm on the other side of it. (a) Name the kind of lens. (b) Find: (i) the focal length of the lens, and (ii) the magnification of the image.
Answer:
(a) Kind of lens
Since the image is formed on a screen, it is a real image. Only a convex lens can form a real image. Thus, the lens is a convex lens.
(b) Focal length and magnification
Given:
Object distance, u = -45 cm
Image distance, v = +90 cm (Image is on the other side)
(i) The focal length of the lens
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/90 – 1/(-45)
=> 1/f = 1/90 + 1/45
=> 1/f = (1 + 2) / 90
=> 1/f = 3 / 90
=> 1/f = 1 / 30
=> f = +30 cm.
Thus, the focal length of the lens is 30 cm.
(ii) The magnification of the image
To find:
Magnification, m = ?
Solution:
Using the magnification formula,
m = v / u
=> m = 90 / (-45)
=> m = -2.
Thus, the magnification of the image is -2.
8. A convex lens forms an inverted image of size same as that of the object which is placed at a distance 60 cm in front of the lens. Find: (a) the position of the image, and (b) the focal length of the lens.
Answer:
Given:
Lens is convex.
Object distance, u = -60 cm
Magnification, m = -1 (Image is inverted and of the same size)
(a) The position of the image
To find:
Image distance, v = ?
Solution:
Using the magnification formula,
m = v / u
=> -1 = v / (-60)
=> v = (-1) × (-60)
=> v = +60 cm.
Thus, the image is formed at a distance of 60 cm behind the lens.
(b) The focal length of the lens
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/60 – 1/(-60)
=> 1/f = 1/60 + 1/60
=> 1/f = 2 / 60
=> 1/f = 1 / 30
=> f = +30 cm.
Thus, the focal length of the lens is 30 cm.
9. A concave lens forms an erect image of 1/3 rd the size of the object which is placed at a distance 30 cm in front of the lens. Find: (a) the position of the image, and (b) the focal length of the lens.
Answer:
Given:
Lens is concave.
Object distance, u = -30 cm
Magnification, m = +1/3 (Image is erect and 1/3rd the size)
(a) The position of the image
To find:
Image distance, v = ?
Solution:
Using the magnification formula,
m = v / u
=> +1/3 = v / (-30)
=> v = (-30) / 3
=> v = -10 cm.
Thus, the image is formed at a distance of 10 cm in front of the lens.
(b) The focal length of the lens
To find:
Focal length, f = ?
Solution:
Using the lens formula,
1/f = 1/v – 1/u
=> 1/f = 1/(-10) – 1/(-30)
=> 1/f = -1/10 + 1/30
=> 1/f = (-3 + 1) / 30
=> 1/f = -2 / 30
=> 1/f = -1 / 15
=> f = -15 cm.
Thus, the focal length of the concave lens is 15 cm.
10. The power of a lens is +2.0 D. Find its focal length and state the kind of the lens.
Answer:
Given:
Power of the lens, P = +2.0 D
To find:
Focal length, f = ?
Kind of the lens = ?
Solution:
Using the power formula,
P = 1 / f (where f is in metres)
=> +2.0 = 1 / f
=> f = 1 / 2.0
=> f = 0.5 m.
In centimetres, f = 0.5 × 100 = 50 cm.
Since the power is positive, the lens is a convex lens.
The focal length is 50 cm.
11. Express the power (with sign) of a concave lens of focal length 20 cm.
Answer:
Given:
Lens is concave.
Focal length = 20 cm.
So, f = -20 cm = -0.20 m.
To find:
Power, P = ?
Solution:
Using the power formula,
P = 1 / f (where f is in metres)
=> P = 1 / (-0.20)
=> P = -5.0 D.
Thus, the power of the concave lens is -5.0 D.
12. The focal length of a convex lens is 25 cm. Express its power with sign.
Answer:
Given:
Lens is convex.
Focal length = 25 cm.
So, f = +25 cm = +0.25 m.
To find:
Power, P = ?
Solution:
Using the power formula,
P = 1 / f (where f is in metres)
=> P = 1 / (+0.25)
=> P = +4.0 D.
Thus, the power of the convex lens is +4.0 D.
13. The power of a lens is -2.0 D. Find its focal length and its kind.
Answer:
Given:
Power of the lens, P = -2.0 D
To find:
Focal length, f = ?
Kind of the lens = ?
Solution:
Using the power formula,
P = 1 / f (where f is in metres)
=> -2.0 = 1 / f
=> f = 1 / (-2.0)
=> f = -0.5 m.
In centimetres, f = -0.5 × 100 = -50 cm.
Since the power is negative, the lens is a concave lens.
The focal length is 50 cm.
14. The magnification by a lens is -3. Name the lens and state how are u and v related?
Answer:
Given:
Magnification, m = -3
To find:
Name of the lens = ?
Relation between u and v = ?
Solution:
Since the magnification is negative, the image formed is real and inverted. A real image can only be formed by a convex lens.
The relation between u and v is given by the magnification formula:
m = v / u
=> -3 = v / u
=> v = -3u
This means the image distance is three times the object distance, and they are on opposite sides of the lens (since u is negative, v will be positive).
15. The magnification by a lens is +0.5. Name the lens and state how are u and v related?
Answer:
Given:
Magnification, m = +0.5
To find:
Name of the lens = ?
Relation between u and v = ?
Solution:
Since the magnification is positive, the image formed is virtual and erect. Since the magnitude of magnification is less than 1 (0.5 < 1), the image is diminished. A virtual and diminished image is always formed by a concave lens.
The relation between u and v is given by the magnification formula:
m = v / u
=> +0.5 = v / u
=> v = 0.5u
This means the image distance is half the object distance, and they are on the same side of the lens (since u is negative, v will also be negative).
16. A photographer needs a lens with focal length of 0.2 m for close up shots. What is the power of the lens required? How does a lens with higher power affect the focussing capability of a camera?
Answer:
Given:
Focal length, f = +0.2 m (A camera uses a convex lens)
To find:
Power, P = ?
Effect of higher power on focusing capability = ?
Solution:
Using the power formula,
P = 1 / f (where f is in metres)
=> P = 1 / 0.2
=> P = +5.0 D.
The power of the lens required is +5.0 D.
A lens with higher power has a shorter focal length (since P = 1/f). A shorter focal length means the lens can bend light rays more strongly. This allows the camera to focus on objects that are closer to it. Therefore, a lens with higher power improves the camera’s ability to focus on nearby objects for close-up shots.
17. A concave lens has a focal length of 30 cm. Find the position and magnification (m) of the image for an object placed in front of it at a distance of 30 cm. State whether the image is real or virtual?
Answer:
Given:
Lens is concave, so focal length, f = -30 cm
Object distance, u = -30 cm
To find:
Image position, v = ?
Magnification, m = ?
Nature of the image = ?
Solution:
Using the lens formula to find v,
1/f = 1/v – 1/u
=> 1/(-30) = 1/v – 1/(-30)
=> -1/30 = 1/v + 1/30
=> 1/v = -1/30 – 1/30
=> 1/v = -2/30
=> 1/v = -1/15
=> v = -15 cm.
The image is formed at a distance of 15 cm in front of the lens.
Now, using the magnification formula,
m = v / u
=> m = (-15) / (-30)
=> m = +0.5.
The magnification is +0.5.
Since the image distance ‘v’ is negative (and magnification ‘m’ is positive), the image is virtual and erect.
18. Find the position and magnification of the image of an object placed at a distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?
Answer:
Given:
Lens is convex, so focal length, f = +10.0 cm
Object distance, u = -8.0 cm
To find:
Image position, v = ?
Magnification, m = ?
Whether image is erect or inverted = ?
Solution:
Using the lens formula to find v,
1/f = 1/v – 1/u
=> 1/10 = 1/v – 1/(-8.0)
=> 1/10 = 1/v + 1/8
=> 1/v = 1/10 – 1/8
=> 1/v = (4 – 5) / 40
=> 1/v = -1 / 40
=> v = -40.0 cm.
The position of the image is 40.0 cm in front of the lens.
Now, using the magnification formula,
m = v / u
=> m = (-40.0) / (-8.0)
=> m = +5.0.
The magnification is +5.0.
Since the magnification ‘m’ is positive, the image is erect.
Exercise D
MCQ
1. The least distance of distinct vision of normal eye is:
(a) 25 mm
(b) 25 cm
(c) 2-5 cm
(d) 20 mm
Answer: (b) 25 cm
2. A magnifying glass forms:
(a) a real and diminished image
(b) a real and magnified image
(c) a virtual and magnified image
(d) a virtual and diminished image
Answer: (c) a virtual and magnified image
3. The maximum magnifying power of a convex lens of focal length 5 cm can be :
(a) 25
(b) 10
(c) 1
(d) 6
Answer: (d) 6
4. The human eye is not able to see an object distinctly if it subtends an angle:
(a) equal to 10′
(b) less than 1′
(c) more than 1′
(d) more than 5′
Answer: (b) less than 1′
5. A simple microscope uses a:
(a) convex lens of short focal length
(b) convex lens of large focal length
(c) concave lens of short focal length
(d) concave lens of large focal length
Answer: (a) convex lens of short focal length
6. Magnifying power of a microscope is given as:
(a) M=f+D
(b) M= D/f
(c) M=f+ D/f
(d) M= 1 + D/f
Answer: (d) M= 1 + D/f
7. A person suffering from long sightedness wears spectacles having a ……….. lens and a person suffering from short signtedness wears spectacles having ………. lens.
(a) convex, convex
(b) concave, concave
(c) concave, convex
(d) convex, concave
Answer: (d) convex, concave
8. A boy has two lenses A and B. When lens A is kept near a printed page, letters appear magnified. When lens B is used to see a distant object, an upright image is seen. The lenses A and B are:
(a) both are convex
(b) both are concave
(c) A→ concave, B → convex
(d) A convex, B→ concave
Answer: (d) A convex, B→ concave
Short Answer Type Questions
1. What is a magnifying glass? State its two uses.
Answer: A magnifying glass, also known as a simple microscope, is a convex lens of short focal length, usually mounted in a lens holder or fitted in a steel or plastic frame provided with a handle.
Two uses of a magnifying glass are:
(i) It is used to see and read small letters and figures.
(ii) It is used by watch-makers to see small parts and screws of a watch.
2. Where is the object placed with respect to the principal focus of a magnifying glass, so as to see its enlarged image? Where is the image obtained?
Answer: To see its enlarged image using a magnifying glass, the object is placed within the focal length of the lens, specifically between its optical centre and its first principal focus (F₁).
The image is obtained on the same side as the object, behind the object, and is formed at the least distance of distinct vision (D).
3. Write an expression for the magnifying power of a simple microscope. How can it be increased?
Answer: The expression for the magnifying power (M) of a simple microscope is given by:
M = 1 + D/f
where f is the focal length of the lens and D is the least distance of distinct vision.
The magnifying power of the microscope can be increased by using a lens of shorter focal length.
4. State two applications each of a convex and a concave lens.
Answer: Two applications of a convex lens are:
(i) The objective lens of a telescope, camera, or slide projector is a convex lens.
(ii) In the collimator of a spectroscope, a convex lens is used for obtaining a parallel beam of light.
Two applications of a concave lens are:
(i) A person suffering from short-sightedness or myopia wears spectacles having a concave lens.
(ii) A concave lens is used as the eye lens in a Galilean telescope to obtain the final erect image of the object.
5. How will you differentiate between a convex and a concave lens by looking at (i) a distant object, (ii) a printed page?
Answer: You can differentiate between a convex and a concave lens by seeing the image formed:
(i) On seeing a distant object through the lens, if its inverted image is seen, the lens is convex, and if an upright image is seen, the lens is concave.
(ii) On keeping the lens near a printed page, if letters appear magnified, the lens is convex, and if the letters appear diminished, the lens is concave.
6. What would be the effect of a change in the colour of incident light from violet to red on the focal length of the lens? Also what is the name given to this kind of a defect of lens?
Answer: If the colour of incident light is changed from violet to red, the focal length of the lens would change because the refractive index of the material of the lens is different for different colours of light.
The name given to this kind of a defect of a lens, where it forms a coloured or blurred image, is chromatic aberration.
Long Answer Type Questions
1. Draw a neat labelled ray diagram to show the formation of an image by a magnifying glass. State three characteristics of the image.
Answer:
The characteristics of the image formed by a magnifying glass are:
(i) The image is virtual.
(ii) The image is erect or upright.
(iii) The image is magnified.
It is formed on the same side and behind the object.
2. Describe in brief how would you determine the approximate focal length of a convex lens.
Answer: To determine the approximate focal length of a convex lens by the distant object method, the principle is that a beam of parallel rays from a distant object incident on a convex lens gets converged in the focal plane of the lens.
The method is as follows: In an open space, against a white (or light coloured) wall, place a metre rule horizontally with its 0 cm end touching the wall and its other end towards the illuminated object which is at a very large distance (such as a tree or an electric pole etc.). Holding the given lens vertically on the metre rule, focus the object on the wall by moving the lens to and fro along the length of the metre rule. Since the light rays incident from a distant object are nearly parallel, so the image of it formed on the wall is almost in the focal plane of the lens. The distance of the lens from the wall on the metre rule directly gives the approximate focal length of the lens.
The figure shows the formation of the image on the wall of a distant tree situated on the left of the lens (which is not shown in the diagram). The approximate focal length of the lens in figure is 48 cm.
3. The diagram in Fig. 5.62 shows the experimental set up for determination of focal length of a lens using a plane mirror.
(i) Draw two rays from the point O of the object to show the formation of image I at O itself.
(ii) What is the size of the image I ?
(iii) State two more characteristics of the image I.
(iv) Name the distance of the object O from the optical centre of the lens.
(v) To what point will the rays return if the mirror is moved away from the lens by a distance equal to the focal length of the lens ?
Answer: (i) When the object pin O lies in the focal plane of lens L, the rays of light starting from O after refraction from the lens, become parallel to its principal axis and therefore they strike normally on the plane mirror M kept behind the lens. Due to normal incidence (i = 0°) on the plane mirror, angle of reflection r = 0°, i.e., the rays are reflected back by the plane mirror on the same path and then they re-enter the lens as a parallel beam. The lens converges these rays in its focal plane, thus forming the image I just above the object pin O.
(ii) The image I is of the same size as that of the object O.
(iii) Two more characteristics of the image I are that it is inverted and real.
(iv) The distance of the object O from the optical centre of the lens (OL) is equal to the focal length of the lens.
(v) The position of plane mirror relative to the lens L (i.e., the distance LM) does not affect the position of the image as long as the rays from the lens fall normally on the plane mirror M. Therefore, if the mirror is moved away from the lens by a distance equal to the focal length of the lens, the rays will still return to the same point O, forming the image I at O, provided the rays from the lens still fall normally on the plane mirror.
4. Describe how you would determine the focal length of a converging lens, using a plane mirror and one pin. Draw a ray diagram to illustrate your answer.
Answer: To determine the exact focal length of a converging lens (convex lens) by an auxillary plane mirror method, given a vertical stand, a plane mirror, a lens and a pin, we proceed as follows:
(i) Place the lens L on a plane mirror MM’ kept on the horizontal surface of the vertical stand and arrange the pin P horizontally in the clamp so that its tip is vertically above the centre O of the lens L as shown in the figure.
(ii) Adjust the height of the pin on the stand till it has no parallax (difference in position) with its inverted image as seen from vertically above the pin. To check for parallax, keep the eye vertically above the tip of the pin P at a distance nearly 25 cm from it and move it sideways. If the tip of the pin and its image shift together, then there is no parallax which implies that the image is in line of the object.
(iii) Measure the distance x of the pin P from the lens and the distance y of the pin from the mirror, using a metre rule and a plumb line. Calculate the average of the two distances. This gives the focal length of the lens, i.e., f = (x+y)/2.
