CBSE Class 10 Science Light Reflection and Refraction Assignment Set 09

This CBSE Class 10 Science assignment gives 25 chapter-wise practice questions with answers and marks on Chapter 9, Light – Reflection and Refraction. It moves from one-mark definitions and quick facts to two-mark differences, three-mark numericals on mirrors, lenses and refractive index, and a five-mark question on ray diagrams of a concave mirror.

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

Question 1. What is the relationship between the refractive index of two media?
Answer:
The refractive index for light going from medium 1 to medium 2 is the reciprocal of the refractive index for light going from medium 2 to medium 1.
n₁₂ = 1/n₂₁, or n₁₂ × n₂₁ = 1.
In simple words: Going one way is the inverse of going the other way.

Teacher's Note:
1. Example: if n of glass with respect to air is 3/2, n of air with respect to glass is 2/3.
2. The product of the two values is always 1.

Question 2. How is the refractive index of a medium related to the speed of light?
Answer:
The refractive index (n) of a medium is the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.
\( n=\dfrac{c}{v} \), where c = 3 × 10⁸ m/s.
In simple words: The slower light travels in a medium, the larger its refractive index.

Teacher's Note:
1. n is always greater than 1 for a material medium.
2. A higher n means an optically denser medium.

Question 3. What is the S.I. unit of refractive index?
Answer:
Refractive index has no unit.
It is the ratio of two speeds (or of the sines of two angles), so the units cancel.
In simple words: It is a plain number.

Teacher's Note:
1. Ratios of like quantities have no unit.
2. Do not write m/s with n.

Question 4. How is the power of a lens related to its focal length?
Answer:
The power of a lens is the reciprocal of its focal length.
\( P=\dfrac{1}{f} \), with f in metres. The unit is the dioptre (D).
In simple words: A shorter focal length means more power.

Teacher's Note:
1. Convert centimetres into metres before using P = 1/f.
2. Convex lens: positive power; concave lens: negative power.

Question 5. Where is the image formed when an object is at large distance from a concave mirror?
Answer:
The image is formed at the principal focus of the mirror. It is real, inverted and highly diminished (point-sized).
In simple words: Far-away objects focus at the focus.

Teacher's Note:
1. Rays from a distant object are nearly parallel.
2. This is how the focal length is found in the laboratory.

Question 6. Why do we see our image in a shining spoon?
Answer:
The shining surface of the spoon is polished and curved, so it behaves like a spherical mirror and reflects light regularly.
The inner (curved-in) surface acts as a concave mirror and the outer surface as a convex mirror, so we see an image of our face.
In simple words: A polished curved surface works like a curved mirror.

Teacher's Note:
1. Polished surfaces give regular reflection.
2. The image in the inner surface may be inverted if the face is far from the spoon.

Question 7. An object is placed in front of a concave mirror between the pole and the focus of the mirror, what is the nature of the image formed by the mirror?
Answer:
The image is virtual, erect and magnified and is formed behind the mirror.
In simple words: Inside the focus a concave mirror acts like a magnifying mirror.

Teacher's Note:
1. This is the position used in shaving and dentist's mirrors.
2. The image cannot be caught on a screen.

Question 8. What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is + 2?
Answer:
A positive magnification means the image is virtual and erect. Since |m| = 2 > 1, it is also twice as large as the object.
In simple words: Positive m means upright and virtual.

Teacher's Note:
1. The sign of m tells real (negative) or virtual (positive).
2. The value of m tells the size relative to the object.

Question 9. Between which two points of a concave mirror should an object be placed to obtain a magnification of – 2?
Answer:
m = −2 means a real, inverted, enlarged image. This happens when the object is placed between F and C.
Check with f = −10 cm: m = −2 gives v = 2u, and \( \dfrac{1}{2u}+\dfrac{1}{u}=-\dfrac{1}{10} \), so u = −15 cm, which lies between F (10 cm) and C (20 cm).
In simple words: A real, enlarged image needs the object between F and C.

Teacher's Note:
1. The negative sign of m shows a real, inverted image.
2. Verify with a quick numerical example.

Section B: Short Answer Questions (2 marks each)

Question 10. (i) What do you mean by reflection of light? (ii) Which surface of a plane mirror reflects most of the light incident on it: The front smooth surface or the back silvered surface? (iii) Define the term angle of incidence.
Answer:
(i) The return of light into the same medium after striking a surface is called reflection of light.
(ii) The back silvered surface reflects most of the light incident on the mirror.
(iii) The angle between the incident ray and the normal at the point of incidence is called the angle of incidence.
In simple words: Reflection is light bouncing back; the silver coating at the back does most of the reflecting.

Teacher's Note:
1. The angle of incidence is measured from the normal, not from the surface.
2. The glass front of a mirror only protects the silvered layer.

Question 11. (i) What is an optical medium? (ii) What do you mean by: (a) an optically denser medium? (b) an optically rarer medium? (iii) Define the term angle of refraction.
Answer:
(i) A medium through which light can travel is called an optical medium.
(ii)(a) While passing from one medium to another, if the speed of light decreases, the second medium is optically denser than the first.
(ii)(b) If the speed of light increases, the second medium is optically rarer than the first.
(iii) The angle between the refracted ray and the normal at the point of incidence is called the angle of refraction.
In simple words: Denser means slower light; rarer means faster light.

Teacher's Note:
1. Optical density is not the same as mass density.
2. Light bends towards the normal in the denser medium.

Question 12. Define the following terms related to spherical mirrors: (i) Centre of curvature (ii) Pole (iii) Aperture
Answer:
(i) The centre of curvature of a spherical mirror is the centre of the sphere of which the mirror is a part.
(ii) The pole is the geometric centre of the reflecting surface of the mirror.
(iii) The aperture is the diameter of the reflecting surface (the area through which light enters and falls on the mirror).
In simple words: C is the centre of the parent sphere, P is the middle of the mirror, and aperture is its width.

Teacher's Note:
1. The radius of curvature is the distance PC.
2. Mirror formulae are valid for mirrors of small aperture.

Question 13. (i) Define: (a) Focus of a concave mirror, (b) Focus of a convex mirror. (ii) Name the mirror that has: (a) a real focus, (b) a virtual focus.
Answer:
(i)(a) The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to the principal axis pass after reflection from the mirror.
(i)(b) The focus of a convex mirror is a point on the principal axis from which the light rays incident parallel to the principal axis appear to diverge after reflection.
(ii)(a) A concave mirror has a real focus.
(ii)(b) A convex mirror has a virtual focus.
In simple words: Concave rays really meet at F; convex rays only seem to come from F.

Teacher's Note:
1. Real focus means the reflected rays actually pass through it.
2. The focal length is half the radius of curvature.

Question 14. (i) What is meant by magnification? Write its expression. (ii) What is its sign for (a) real, (b) virtual image?
Answer:
(i) Magnification is the ratio of the height of the image to the height of the object: \( m=\dfrac{h'}{h}=-\dfrac{v}{u} \).
(ii)(a) For a real image the magnification is negative (the image is inverted).
(ii)(b) For a virtual image the magnification is positive (the image is erect).
In simple words: m tells how big the image is and whether it is upright or upside down.

Teacher's Note:
1. Heights above the axis are positive and below the axis are negative.
2. |m| > 1 enlarged, |m| < 1 diminished, |m| = 1 same size.

Teacher's Note:
1. The source describes concave by silvering the outer surface; remember the reflecting surface is the curved-in one.
2. Give at least three points for 3 marks.

Teacher's Note:
1. Sign of magnification: real is negative, virtual is positive.
2. A plane mirror always forms a virtual image.

Teacher's Note:
1. Convex lens is also called a converging lens and concave lens a diverging lens.
2. Remember the sign of focal length for each.

Section C: Numerical Questions (3 marks each)

Question 18. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
R = +32 cm for a convex mirror.
f = R/2 = 32/2 = +16 cm.
In simple words: The focal length is half the radius of curvature, 16 cm.

Teacher's Note:
1. A convex mirror has a positive focal length.
2. Use f = R/2.

Question 19. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
f = +15 cm, u = −10 cm.
\( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{15}+\dfrac{1}{10}=\dfrac{5}{30}=\dfrac{1}{6} \), so v = +6 cm.
The image is formed 6 cm behind the mirror. m = −v/u = +0.6, so it is virtual, erect and diminished.
In simple words: The image is 6 cm behind the mirror, upright and smaller.

Teacher's Note:
1. Positive v means behind a mirror.
2. m = +0.6 shows an erect image 0.6 times the object.

Question 20. The absolute refractive index of Ruby is 1.7. Find the speed of light in Ruby. The speed of light in vacuum is 3 × 10⁸ m/s.
Answer:
n = c/v, so v = c/n = (3 × 10⁸)/1.7.
v = 1.76 × 10⁸ m/s (approximately).
In simple words: Divide the speed in vacuum by 1.7.

Teacher's Note:
1. Speed in a medium is always less than c.
2. Keep the powers of ten together.

Question 21. Find the power of a concave lens of focal length 2 m.
Answer:
For a concave lens f = −2 m.
P = 1/f = 1/(−2) = −0.5 D.
In simple words: A concave lens has negative power, here −0.5 D.

Teacher's Note:
1. f is already in metres.
2. The negative sign shows a diverging lens.

Question 22. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
u = −25 cm, f = +10 cm, h = 5 cm.
\( \dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{10}-\dfrac{1}{25}=\dfrac{3}{50} \), so v = +16.67 cm (on the other side of the lens).
m = v/u = 16.67/(−25) = −0.67, and h′ = −0.67 × 5 ≈ −3.3 cm.
The image is real, inverted and diminished (3.3 cm tall), formed between F₂ and 2F₂.

In simple words: The image is 16.67 cm away, upside down and about two-thirds of the object's size.

Teacher's Note:
1. The object is beyond 2F, so the image lies between F and 2F on the other side.
2. Negative image height means an inverted image.

Question 23. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
f = −15 cm, v = −10 cm (virtual image on the same side as the object).
\( \dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{10}+\dfrac{1}{15}=-\dfrac{1}{30} \), so u = −30 cm.
The object is 30 cm in front of the lens. The magnification is m = v/u = +1/3, so the image is virtual, erect and diminished.

In simple words: The object is 30 cm from the lens, and the image is one-third its size.

Teacher's Note:
1. For a concave lens the image is always virtual and on the same side as the object.
2. A negative u means the object is in front of the lens.

Question 24. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image formed.
Answer:
h = +10 cm, f = +12 cm, u = −18 cm.
\( \dfrac{1}{v}=\dfrac{1}{12}-\dfrac{1}{18}=\dfrac{3-2}{36}=\dfrac{1}{36} \), so v = +36 cm.
m = v/u = 36/(−18) = −2, so h′ = −2 × 10 = −20 cm.
The image is real, inverted and enlarged, 20 cm tall, formed 36 cm from the lens on the other side.
In simple words: The image is 36 cm away, upside down and twice as tall (20 cm).

Teacher's Note:
1. The object lies between F and 2F, so the image is beyond 2F.
2. Check: m = −2 matches the enlarged real image.

Section D: Long Answer Question (5 marks)

Question 25. Draw the ray diagrams showing the image formation by a concave mirror when an object is placed. (i) Between the pole and focus of the mirror. (ii) Between focus and centre of curvature of the mirror. (iii) At centre of curvature of the mirror. (iv) A little beyond centre of curvature of the mirror. (v) At infinity.
Answer:
Diagrams below are drawn for a concave mirror of focal length 10 cm.
(i) Object between P and F (u = −5 cm): image behind the mirror, virtual, erect and magnified.

(ii) Object between F and C (u = −15 cm): image beyond C, real, inverted and magnified.

(iii) Object at C (u = −20 cm): image at C, real, inverted and of the same size.

(iv) Object a little beyond C (u = −30 cm): image between F and C, real, inverted and diminished.

(v) Object at infinity: image at F, real, inverted and highly diminished (point-sized).

In simple words: As the object moves from infinity towards the mirror, the real image moves out from F to infinity and grows; inside F it turns virtual.

Teacher's Note:
1. Use any two of the standard rays: parallel to the axis, through F, through C, or striking the pole.
2. Mark P, F and C and the arrow heads on the rays.