CBSE Class 10 Science Light reflection and refraction Sure Shot Questions Set 03

This CBSE Class 10 Science page lists 21 sure shot questions with answers from Chapter 9, Light – Reflection and Refraction, picked from the questions that appear most often in board papers. It covers laws of reflection and refraction, image formation by concave and convex mirrors and lenses, refractive index, power of a lens and the numericals with ray diagrams, each solved step by step.

Very Short Answer Questions

Question 1. How is the focal length of a spherical mirror related to its radius of curvature?
Answer:
The focal length of a spherical mirror is half of its radius of curvature.
\( f=\dfrac{R}{2} \), or R = 2f.
In simple words: The focus lies midway between the pole and the centre of curvature.

Teacher's Note:
1. This relation is valid for mirrors of small aperture only.
2. In numericals, find f first if R is given, and take its sign from the Cartesian convention.

Question 2. At what position the object be placed in front of a concave mirror to form a real image of the same size?
Answer:
The object must be placed at the centre of curvature (C) of the concave mirror.
The image is then formed at C itself, and it is real, inverted and of the same size (m = −1).
In simple words: Place the object at C and the image also appears at C, equal in size.

Teacher's Note:
1. At C the object distance equals 2f.
2. Remember m = −1 for this case.

Question 3. A ray of light moving along the principal axis is falling on a concave mirror. In which direction is it reflected?
Answer:
The ray is reflected back along the principal axis itself.
The principal axis passes through the centre of curvature, so the ray strikes the mirror along the normal. Hence ∠i = ∠r = 0°.
In simple words: A ray that hits the mirror along the normal comes straight back.

Teacher's Note:
1. The line joining pole and centre of curvature is always normal to the mirror at the pole.
2. The same reasoning applies to any ray passing through the centre of curvature.

Question 4. For the same angle of incidence of 45°, the refractive angle in two transparent media P and Q is 20° and 30°, respectively. Which medium is optically denser out of P and Q and why?
Answer:
Refractive index = sin i / sin r.
For medium P: n = sin 45° / sin 20° = 0.707 / 0.342 = 2.07.
For medium Q: n = sin 45° / sin 30° = 0.707 / 0.5 = 1.41.
Medium P has the larger refractive index, so medium P is optically denser. The ray bends more towards the normal in P (smaller angle of refraction).
In simple words: The medium that bends the ray closer to the normal is the denser one.

Teacher's Note:
1. For the same angle of incidence, a smaller angle of refraction means a larger refractive index.
2. A denser medium means a slower speed of light in it.

Question 5. Name the type of lens used to obtain (i) an erect, enlarged and virtual image of an object, (ii) an erect, diminished and virtual image of an object.
Answer:
(i) A convex lens, with the object placed between the optical centre and the principal focus.
(ii) A concave lens, which gives an erect, diminished, virtual image for every position of the object.
In simple words: Convex lens magnifies only when used as a magnifier; a concave lens always shrinks.

Teacher's Note:
1. A convex lens can give both real and virtual images; a concave lens gives only virtual images.
2. Link the nature of the image to the position of the object.

Short Answer Questions

Question 6. (i) State the two laws of reflection of light. (ii) A light ray is incident normally on a plane mirror. What is its angle of incidence?
Answer:
(i) First law: The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane.
Second law: The angle of incidence is equal to the angle of reflection (∠i = ∠r).
(ii) A ray incident normally falls along the normal, so the angle of incidence is 0°.
In simple words: The reflected ray lies in the same plane and leaves at the same angle at which it came.

Teacher's Note:
1. Angles are always measured from the normal, not from the mirror surface.
2. These laws hold for plane as well as curved reflecting surfaces.

Question 7. (i) State the two laws of refraction of light. (ii) A light ray is incident normally on a glass block. What will be the angle of refraction?
Answer:
(i) First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
Second law (Snell's law): For light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant: \( \dfrac{\sin i}{\sin r}=\text{constant}=n \). This constant is the refractive index of the second medium with respect to the first.
(ii) A normally incident ray does not bend, so the angle of refraction is 0°.
In simple words: Light bends at the boundary in a fixed way, and a ray along the normal goes straight.

Teacher's Note:
1. The constant in Snell's law depends on the two media and on the colour of light.
2. For i = 0° we get r = 0° because sin 0° = 0.

Question 8. (i) What is relative refractive index? (ii) Define one dioptre of power of a lens. (iii) How lateral shift depends on the angle of incidence?
Answer:
(i) The refractive index of one medium with respect to another medium is called the relative refractive index.
(ii) One dioptre is the power of a lens whose focal length is 1 metre.
(iii) The lateral shift of a ray passing through a glass slab increases as the angle of incidence increases (it is zero for normal incidence).
In simple words: A bigger slant at entry gives a bigger sideways shift.

Teacher's Note:
1. The shift is only roughly proportional to the angle of incidence for small angles, so write that it increases with the angle of incidence.
2. Lateral shift also increases with the thickness of the slab.

Question 9. The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. If speed of light in glass is 2 × 10⁸ m/s, find the speed of light in water.
Answer:
Refractive index = speed of light in air ÷ speed of light in the medium.
For glass: \( \dfrac{3}{2}=\dfrac{c}{2\times10^{8}} \), so c = 3 × 10⁸ m/s.
For water: \( \dfrac{4}{3}=\dfrac{3\times10^{8}}{v_w} \), so v_w = \( \dfrac{9}{4}\times10^{8} \) = 2.25 × 10⁸ m/s.
In simple words: Find the speed in air from glass first, then divide it by the refractive index of water.

Teacher's Note:
1. Use the speed of light in air (3 × 10⁸ m/s) as the common reference.
2. The denser the medium, the smaller the speed of light in it.

Question 10. Explain why a ray of light passing through the centre of curvature of a concave mirror gets reflected among the same path after reflection.
Answer:
A ray passing through the centre of curvature falls on the mirror along the radius, which is the normal to the mirror at that point.
So the angle of incidence is 0°, and by the law of reflection the angle of reflection is also 0°. The ray is therefore reflected back along the same path.
In simple words: The ray hits the mirror along the normal, so it comes straight back.

Teacher's Note:
1. Radius of a spherical surface is always normal to the surface.
2. Use this ray as one of the standard rays in ray diagrams.

Question 11. Why are concave mirrors used in a solar furnace?
Answer:
A concave mirror is a converging mirror. The parallel rays of the Sun fall on it and, after reflection, all of them pass through the principal focus.
When the furnace is placed at the focus, a large amount of the Sun's energy is concentrated there, so the temperature becomes very high.
In simple words: The mirror collects the Sun's rays on one small spot and makes it very hot.

Teacher's Note:
1. Large aperture mirrors collect more sunlight and give higher temperature.
2. The same idea is used in solar cookers.

Question 12. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes, the lens still produces a complete image of the object.
Rays from every point of the object reach the lens, and the uncovered half still refracts them to meet at the image point. So each point of the object has its image, only the number of rays is halved.
Experiment: Form a sharp image of a distant object on a screen with the full lens, then cover one half with black paper. The image remains complete, but it becomes less bright (its intensity decreases).
In simple words: A half-covered lens still gives the full image, only dimmer.

Teacher's Note:
1. Each part of a lens can form the whole image.
2. Mention the fall in brightness of the image.

Long Answer Questions

Question 13. An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.
Answer:
Given: h = +4 cm, u = −15 cm, f = −10 cm.
Mirror formula: \( \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \), so \( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{1}{10}+\dfrac{1}{15}=-\dfrac{1}{30} \), giving v = −30 cm.
The screen must be placed 30 cm in front of the mirror (on the object side).
Magnification m = −v/u = −(−30)/(−15) = −2.
Height of image h′ = m × h = −2 × 4 = −8 cm. The image is 8 cm tall, real and inverted.
In simple words: The screen goes 30 cm from the mirror and the image is twice the object, 8 cm, upside down.

Teacher's Note:
1. Substitute u and f with their negative signs for a concave mirror.
2. A real image can be caught on a screen; its height is negative (inverted).

Question 14. The image formed by a spherical mirror is real, inverted and is of magnification – 2. If the image is at a distance of 30 cm from the mirror, where is the object placed? Find the focal length of the mirror. List two characteristics of the image formed if the object is moved 10 cm towards the mirror.
Answer:
Given: m = −2, v = −30 cm (real image is in front of the mirror).
m = −v/u gives −2 = −(−30)/u, so u = −15 cm. The object is 15 cm in front of the mirror.
\( \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{1}{30}-\dfrac{1}{15}=-\dfrac{1}{10} \), so f = −10 cm (concave mirror).
If the object is moved 10 cm towards the mirror, u = −5 cm, which lies between P and F (5 cm < 10 cm). The image formed is virtual and erect (also magnified, behind the mirror).
In simple words: The object is at 15 cm, the focal length is 10 cm, and moving closer than the focus gives a virtual, upright image.

Teacher's Note:
1. Compare the new object distance with f to decide the type of image.
2. Between P and F a concave mirror always gives a virtual, erect, enlarged image.

Question 15. A 6 cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30 cm. The distance of the object from the mirror is 45 cm. Use mirror formula to determine the position, nature and size of the image formed. Also draw labelled ray diagram to show the image formation in this case.
Answer:
Given: h = +6 cm, f = −30 cm, u = −45 cm.
\( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{1}{30}+\dfrac{1}{45}=-\dfrac{1}{90} \), so v = −90 cm: the image is 90 cm in front of the mirror, beyond C (C is at 60 cm).
m = −v/u = −(−90)/(−45) = −2, so h′ = −2 × 6 = −12 cm.
The image is real, inverted and enlarged (12 cm tall).

In simple words: The image forms 90 cm away, twice the size of the object, real and upside down.

Teacher's Note:
1. C is at 60 cm and F at 30 cm, so an object at 45 cm lies between F and C.
2. For this position the image is real, inverted, enlarged and beyond C.

Question 16. An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. (i) Use lens formula to find the distance of the image from the lens. (ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case. (iii) Draw ray diagram to justify your answer of part (ii).
Answer:
Given: u = −60 cm, f = −30 cm (concave lens).
(i) \( \dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=-\dfrac{1}{30}-\dfrac{1}{60}=-\dfrac{3}{60}=-\dfrac{1}{20} \), so v = −20 cm.
(ii) m = v/u = (−20)/(−60) = +1/3. The image is virtual, erect, diminished (one-third of the object) and formed on the same side as the object, 20 cm from the lens, between the optical centre and the focus F₁.
(iii) Ray diagram:

In simple words: A concave lens makes a small, upright, virtual image between the lens and its focus.

Teacher's Note:
1. Use the lens formula with f negative for a concave lens.
2. m = +1/3 means erect and one-third the size.

Question 17. (i) Define focal length of a divergent lens. (ii) A divergent lens of focal length 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed. (iii) Draw a ray diagram to show the formation of image in the above situation.
Answer:
(i) The distance between the optical centre and the point from which the parallel rays, after refraction through a divergent lens, appear to diverge is called its focal length.
(ii) Given: f = −30 cm, v = −15 cm, h = 6 cm.
\( \dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{15}+\dfrac{1}{30}=-\dfrac{1}{30} \), so u = −30 cm.
m = v/u = (−15)/(−30) = +0.5, so image size h′ = 0.5 × 6 = 3 cm (erect, virtual).
(iii) Ray diagram:

In simple words: The object is 30 cm from the lens and its image is half its size, 3 cm.

Teacher's Note:
1. For a diverging lens, v and f are both negative for a real object.
2. Image height is positive, so the image is erect.

Question 18. (i) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual, state the type of the mirror and also draw a ray diagram to justify your answer. Write one use of such mirrors are put to and why? (ii) Define the radius of curvature of spherical mirrors. Find the nature and focal length of a spherical mirror whose radius of curvature is + 24 cm.
Answer:
(i) The mirror is a convex mirror. It always forms a virtual, erect and diminished image behind the mirror, between P and F.

Use: Convex mirrors are used as rear-view (side) mirrors in vehicles because they give an erect, diminished image and a wide field of view, so the driver sees a large area behind.
(ii) The radius of curvature is the distance between the pole and the centre of curvature of a spherical mirror (the radius of the sphere of which the mirror is a part).
R = +24 cm (positive), so the mirror is convex and f = R/2 = +12 cm.
In simple words: Only a convex mirror always gives a small, upright, virtual image, which is why it is used for rear views.

Teacher's Note:
1. A positive focal length means a convex mirror in the Cartesian sign convention.
2. Mention 'wider field of view' when asked why convex mirrors are used.

Question 19. It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm. (i) What should be the range of distance of an object placed in front of the mirror? (ii) Will the image be smaller or larger than the object? Draw ray diagram to show the formation of image in this case. (iii) Where will the image of this object be, if it is placed 24 cm in front of the mirror? Draw ray diagram for this situation also to justify your answer. Show the positions of pole, principal focus and the centre of curvature in the above ray diagrams.
Answer:
(i) A concave mirror forms an erect image only when the object is between the pole and the focus, so the object distance should lie between 0 and 12 cm.
(ii) The image is virtual, erect and larger than the object, formed behind the mirror. Diagram (object at 6 cm):

(iii) For u = −24 cm and f = −12 cm: \( \dfrac{1}{v}=-\dfrac{1}{12}+\dfrac{1}{24}=-\dfrac{1}{24} \), so v = −24 cm. The image forms 24 cm in front of the mirror, at C; it is real, inverted and of the same size. Diagram:

In simple words: Inside the focus the image is large and upright; at 24 cm the image is at the same place, same size, upside down.

Teacher's Note:
1. The centre of curvature is at 2f = 24 cm, so the object at 24 cm is at C.
2. Show P, F and C on both diagrams, as the question asks.

Teacher's Note:
1. Only observation 6 breaks the rule that an object within f gives a virtual image.
2. Check the other rows with the lens formula: 1/v − 1/u = 1/20.

Question 21. An object 1 m tall is placed on the principal axis of a convex lens and its 40 cm tall image is formed on the screen placed at a distance of 70 cm from the object. What is the focal length of the lens?
Answer:
The image is on a screen, so it is real and inverted: h = 100 cm, h′ = −40 cm, so m = −0.4.
Let the object be x cm from the lens. Then v = 70 − x and m = v/u = (70 − x)/(−x) = −0.4, so 70 − x = 0.4x and x = 50 cm.
Hence u = −50 cm and v = +20 cm.
\( \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{20}+\dfrac{1}{50}=\dfrac{7}{100} \), so f = 100/7 = +14.3 cm (approx.).
In simple words: The object is 50 cm and the screen 20 cm from the lens, giving a focal length of about 14.3 cm.

Teacher's Note:
1. The object-to-screen distance is |u| + v = 70 cm.
2. The negative magnification matches a real, inverted image.