CBSE Class 10 Science Light Reflection and Refraction Assignment Set 04

This CBSE Class 10 Science assignment contains 25 practice questions with answers and marks from Chapter 9, Light – Reflection and Refraction. It revises plane mirror images, convex and concave lenses, power and refractive index through short questions, three-mark numericals with ray diagrams, and five-mark questions on image formation by a convex lens.

Section A: Very Short Answer Questions (1 mark each)

Question 1. Define refraction.
Answer:
The change in the direction of a ray of light when it passes obliquely from one transparent medium into another is called refraction of light.
It happens because the speed of light is different in different media.
In simple words: Refraction is the bending of light at the boundary between two media.

Teacher's Note:
1. Mention 'oblique' incidence, as a ray along the normal does not bend.
2. The cause is the change in speed of light.

Question 2. Define refractive index.
Answer:
The refractive index of a medium is the ratio of the speed of light in vacuum (air) to the speed of light in that medium: \( n=\dfrac{c}{v} \).
For a ray going from medium 1 to medium 2 it is also the ratio \( \dfrac{\sin i}{\sin r} \) of the sine of the angle of incidence to the sine of the angle of refraction.
In simple words: It tells how many times slower light travels in the medium than in vacuum.

Teacher's Note:
1. The refractive index has no unit.
2. A larger n means the ray bends more.

Question 3. What is the angle of incidence if a ray of light is incident normal to the surface separating the two media?
Answer:
The angle of incidence is 0°, because the ray travels along the normal. The angle of refraction is also 0°, so the ray passes undeviated.
In simple words: A ray along the normal has zero angle of incidence.

Teacher's Note:
1. The angle is measured from the normal, not from the surface.
2. No bending occurs for normal incidence.

Question 4. A thin lens has a focal length f = −12 cm. Is it convex or concave lens?
Answer:
The focal length is negative, so it is a concave lens (a diverging lens).
In simple words: Negative focal length means a concave lens.

Teacher's Note:
1. Convex lens: f positive. Concave lens: f negative.
2. Its power would be −1/0.12 ≈ −8.3 D.

Question 5. A lens forms an erect image for all positions of the object in front of it. Is the lens convex or concave?
Answer:
The lens is concave.
A concave lens always forms a virtual, erect and diminished image. A convex lens forms an erect image only when the object is within its focal length, not for all positions.
In simple words: Only a concave lens gives an erect image wherever the object is.

Teacher's Note:
1. The words 'for all positions' decide the answer.
2. A convex lens gives a real, inverted image beyond its focus.

Question 6. Where should an object be placed so that a real and inverted image of same size is obtained using a convex lens?
Answer:
The object should be placed at 2F₁ (at a distance 2f from the lens). The image is then formed at 2F₂ on the other side, real, inverted and of the same size.
In simple words: Place the object at twice the focal length.

Teacher's Note:
1. At 2F the object and image distances are equal.
2. Magnification is −1.

Question 7. What happens when a ray of light passes through the optical centre of a lens?
Answer:
A ray passing through the optical centre of a lens goes straight without any deviation, because there the two faces of a thin lens are almost parallel.
In simple words: The ray through the centre of a lens does not bend.

Teacher's Note:
1. This is one of the standard rays used in ray diagrams.
2. It is true for both convex and concave lenses.

Question 8. Name the mirror that a shopkeeper can use to check theft in his shops.
Answer:
A convex mirror. It has a wide field of view, so the shopkeeper can watch a large part of the shop in one small mirror.
In simple words: A convex mirror shows a wide area.

Teacher's Note:
1. The same reason makes convex mirrors suitable for rear-view mirrors.
2. The image is small, erect and virtual.

Section B: Short Answer Questions (2 to 3 marks each)

Question 9. List out the factors on which the refractive index of a medium depends.
Answer:
The refractive index of a medium depends on:
1. the nature (material) of the medium, i.e. its optical density;
2. the colour (wavelength) of the light used;
3. the temperature of the medium.
In simple words: Refractive index depends on the material, the colour of light and the temperature.

Teacher's Note:
1. It does not depend on the angle of incidence.
2. This is why light of a single colour is used when measuring it.

Question 10. An object is kept in front of a plane mirror. If the mirror is moved away from the object through a distance x, by what distance will the image move?
Answer:
In a plane mirror the image is as far behind the mirror as the object is in front.
If the mirror moves away from the object by x, the object–mirror distance increases by x and the mirror itself also shifts by x, so the image moves by x + x = 2x away from the object.
In simple words: The image shifts twice as much as the mirror does.

Teacher's Note:
1. Take the position of the mirror as the reference and add the two shifts.
2. Image distance always equals object distance.

Question 11. An insect is at a distance of 1.5 m from a plane mirror. Calculate the distance: (i) At which the image of the insect is formed. (ii) Between the insect and its image.
Answer:
(i) The image in a plane mirror is as far behind the mirror as the object is in front. So the image is formed 1.5 m behind the mirror.
(ii) The distance between the insect and its image = 1.5 + 1.5 = 3 m.
In simple words: The image is 1.5 m behind the mirror, so the insect and image are 3 m apart.

Teacher's Note:
1. The image distance equals the object distance for a plane mirror.
2. Add the two distances for the separation.

Question 12. A concave mirror is cut out of a hollow glass sphere of radius 30 cm. Calculate the focal length of the mirror.
Answer:
The radius of curvature of the mirror is the radius of the sphere, R = 30 cm.
f = R/2 = 15 cm. For a concave mirror it is written as f = −15 cm in the Cartesian sign convention.
In simple words: The focal length is half of 30 cm, i.e. 15 cm.

Teacher's Note:
1. The mirror takes the radius of curvature of the sphere it is cut from.
2. Use f = R/2.

Question 13. List four characteristics of the images formed by plane mirrors.
Answer:
The image formed by a plane mirror is:
1. virtual (it cannot be obtained on a screen);
2. erect;
3. of the same size as the object;
4. formed as far behind the mirror as the object is in front, and laterally inverted.
In simple words: A plane mirror gives an upright, same-size virtual image behind the mirror.

Teacher's Note:
1. Any four properties are enough.
2. Lateral inversion means left and right are interchanged.

Question 14. What is the magnification of the images formed by plane mirrors and why?
Answer:
The magnification of a plane mirror is +1.
The image is of the same size as the object (h′ = h) and is erect, so m = h′/h = +1. Also v = −u, so m = −v/u = +1.
In simple words: The plus sign means upright and the value 1 means the same size.

Teacher's Note:
1. The positive sign shows a virtual, erect image.
2. This value is the same for every position of the object.

Question 15. What is meant by power of a lens? What does its sign (+ve or –ve) indicate? State its S.I. unit. How is this unit related to focal length of a lens.
Answer:
The power of a lens is the reciprocal of its focal length in metres: \( P=\dfrac{1}{f} \). It tells how strongly the lens converges or diverges light.
Sign: a positive power indicates a convex (converging) lens and a negative power indicates a concave (diverging) lens.
S.I. unit: dioptre (D). One dioptre is the power of a lens whose focal length is 1 metre: 1 D = 1 m⁻¹.
In simple words: Power is 1/f in metres, measured in dioptres, positive for convex and negative for concave lenses.

Teacher's Note:
1. Always convert focal length into metres.
2. Shorter focal length means a greater power.

Section C: Numerical Questions (3 marks each)

Question 16. An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.
Answer:
f = −20 cm, u = −50 cm.
\( \dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=-\dfrac{1}{20}-\dfrac{1}{50}=-\dfrac{7}{100} \), so v = −14.3 cm.
The image is formed 14.3 cm from the lens on the same side as the object. m = v/u = +0.29, so it is virtual, erect and diminished.
In simple words: A concave lens gives a small, upright, virtual image about 14.3 cm from the lens.

Teacher's Note:
1. For a concave lens both v and f are negative for a real object.
2. The image lies between the lens and its focus.

Question 17. An object is placed 50 cm from a lens which produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens.
Answer:
u = −50 cm and v = −10 cm (a virtual image on the object side).
\( \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{10}+\dfrac{1}{50}=-\dfrac{4}{50} \), so f = −12.5 cm.
The focal length is negative, so the lens is concave. Diagram:

In simple words: A virtual diminished image on the object side means a concave lens of focal length 12.5 cm.

Teacher's Note:
1. A virtual image on the same side of the lens as the object has negative v.
2. The image is nearer the lens than the object.

Question 18. An object of height 4 cm is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.
Answer:
h = 4 cm, u = −10 cm, f = +20 cm.
\( \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=-\dfrac{1}{20} \), so v = −20 cm (on the same side as the object).
m = v/u = (−20)/(−10) = +2, so h′ = 2 × 4 = 8 cm.
The image is virtual, erect and enlarged (8 cm tall), 20 cm from the lens on the object side.

In simple words: Inside the focus a convex lens acts as a magnifier: the image is 20 cm away, upright and 8 cm tall.

Teacher's Note:
1. The object is between O and F, so the image is virtual and magnified.
2. Positive m means an erect image.

Question 19. An object is placed in front of a concave mirror of radius of curvature 15 cm at a distance of (a) 10 cm and (b) 5 cm. Find the position, nature and magnification of the image in each case.
Answer:
R = −15 cm, so f = R/2 = −7.5 cm.
(a) u = −10 cm: \( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{1}{7.5}+\dfrac{1}{10}=-\dfrac{1}{30} \), so v = −30 cm. m = −v/u = −3. The image is 30 cm in front of the mirror, real, inverted and enlarged (3 times).
(b) u = −5 cm: \( \dfrac{1}{v}=-\dfrac{1}{7.5}+\dfrac{1}{5}=\dfrac{1}{15} \), so v = +15 cm. m = −v/u = +3. The image is 15 cm behind the mirror, virtual, erect and enlarged (3 times).
In simple words: At 10 cm the image is real and inverted; inside the focus at 5 cm it is virtual and upright, both three times larger.

Teacher's Note:
1. Compare u with f (7.5 cm) to know which type of image to expect.
2. The sign of m tells real or virtual.

Question 20. An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Give the position, magnification and the nature of the image formed.
Answer:
f = +10 cm, u = −5 cm.
\( \dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{5}=\dfrac{3}{10} \), so v = +3.33 cm (behind the mirror).
m = −v/u = −3.33/(−5) = +0.67.
The image is virtual, erect and diminished.
In simple words: The image is 3.33 cm behind the mirror, upright and two-thirds the size.

Teacher's Note:
1. The magnification of a convex mirror is always between 0 and +1.
2. Positive v means behind the mirror.

Question 21. Light travel from rarer medium 1 to a denser medium 2. The angle of incidence and refraction are respectively 45° and 30°. Calculate the refractive index of second medium with respect to the first medium.
Answer:
n₂₁ = sin i / sin r = sin 45° / sin 30°
= 0.7071 / 0.5 = 1.41 (that is, √2).
In simple words: Divide sin 45° by sin 30° to get 1.41.

Teacher's Note:
1. The ray bends towards the normal (r < i), so n > 1, which matches the answer.
2. Refractive index has no unit.

Question 22. With respect to air the refractive index of ice and rock salt are 1.31 and 1.54 respectively. Calculate the refractive index of rock salt with respect to ice.
Answer:
n_{rock salt, air} = 1.54 and n_{ice, air} = 1.31.
n_{rock salt, ice} = n_{rock salt, air} / n_{ice, air} = 1.54/1.31 = 1.18 (approximately).
In simple words: Divide the refractive index of rock salt by that of ice.

Teacher's Note:
1. Both indices are with respect to the same medium (air).
2. Only the refractive indices of ice and rock salt are needed here.

Question 23. A real image 2/3 of the size of an object is formed by a convex lens when the object is at a distance of 12 cm from it. Find the focal length of the lens.
Answer:
The image is real and inverted, so m = −2/3. Also u = −12 cm.
m = v/u gives −2/3 = v/(−12), so v = +8 cm.
\( \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{8}+\dfrac{1}{12}=\dfrac{5}{24} \), so f = +4.8 cm.
In simple words: The image distance is 8 cm and the focal length comes out as 4.8 cm.

Teacher's Note:
1. A real image means m is negative.
2. Check: object beyond 2f (9.6 cm) gives a diminished image.

Section D: Long Answer Questions (5 marks each)

Question 24. Draw ray diagrams showing the image formation by a convex lens when an object is placed (a) between optical centre and focus of the lens (b) between focus and twice the focal length of the lens (c) at twice the focal length of the lens (d) at infinity (e) at the focus of the lens
Answer:
Diagrams for a convex lens of focal length 10 cm:
(a) Between O and F (u = −5 cm): virtual, erect, magnified image on the same side as the object.

(b) Between F and 2F (u = −15 cm): real, inverted, enlarged image beyond 2F.

(c) At 2F (u = −20 cm): real, inverted image of the same size at 2F on the other side.

(d) At infinity: real, inverted, highly diminished (point-sized) image at the focus F₂.

(e) At the focus F₁: the refracted rays are parallel, so the image is formed at infinity; it is real, inverted and highly enlarged.

In simple words: The image moves from F to infinity as the object moves from infinity to F, and becomes virtual once the object is inside F.

Teacher's Note:
1. Use a ray parallel to the axis and a ray through the optical centre.
2. Label F₁, F₂, 2F₁, 2F₂ and O.

Question 25. A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under: Position of candle = 12.0 cm, Position of convex lens = 50.0 cm, Position of the screen = 88.0 cm. (i) What is the focal length of the convex lens? (ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm? (iii) What will be the nature of the image formed if he further shifts the candle towards the lens? (iv) Draw a ray diagram to show the formation of the image in case (iii) as said above.
Answer:
(i) Object distance u = 12 − 50 = −38 cm and image distance v = 88 − 50 = +38 cm. The two are equal, so the object is at 2f: 2f = 38 cm and f = 19 cm.
(ii) With the candle at 31 cm, u = 31 − 50 = −19 cm, which equals the focal length. The object is at F, so the image is formed at infinity and cannot be obtained on the screen.
(iii) If the candle is moved still closer (u less than 19 cm), the object lies between the optical centre and F. The image is virtual, erect and magnified, on the same side as the candle.
(iv) Ray diagram (for example, candle 10 cm from the lens):

In simple words: Equal distances give f = 19 cm; at 19 cm the image goes to infinity; closer than that it becomes a magnified virtual image.

Teacher's Note:
1. Subtract the lens position from the other positions to get u and v.
2. An image formed on a screen is always real.