CBSE Class 10 Physics Light Reflection and Refraction Previous Year Questions

(a) When light travels from a rarer to a denser medium, it bends towards the normal. Since $n_B > n_A$, the ray bends towards the normal when travelling from A to B.

(b) The speed of light increases when it travels from a denser to a rarer medium. Since $n_C < n_B$, the speed of light increases when travelling from B to C.

Reason: Using $v = \dfrac{c}{n}$, a lower refractive index means a higher speed of light.

The completed and labelled ray diagram for refraction through a glass prism is shown below:

Given $P = +5\text{D}$: $$f = \frac{1}{P} = \frac{1}{5} \, \text{m} = 20 \, \text{cm}$$ Since power is positive, the lens is a convex (converging) lens.

To obtain an inverted image of the same size, the object must be placed at 2f from the lens: $$2f = 2 \times 20 = 40 \, \text{cm}$$ Final Answer: Convex lens; focal length = 20 cm; object distance = 40 cm.

$$\text{Speed in X} > \text{Speed in Y} > \text{Speed in Z}$$ Justification: By Snell’s law, $n = \dfrac{\sin i}{\sin r}$. For the same angle of refraction $r$, a block with a smaller angle of incidence has a smaller refractive index.

Block X has the smallest $n$ → maximum speed. Block Z has the largest $n$ → minimum speed. Since $v = \dfrac{c}{n}$, higher refractive index means lower speed.

Given: $h_0 = 10$ cm, $u = -25$ cm, $f = +15$ cm

Step 1 — Image Distance: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{15} – \frac{1}{25} = \frac{5-3}{75} = \frac{2}{75}$$ $$v = +37.5 \, \text{cm}$$ (Positive sign → real image on opposite side of lens.)

Step 2 — Height of Image: $$h_i = \frac{v}{u} \times h_0 = \frac{37.5}{-25} \times 10 = -15 \, \text{cm}$$ (Negative sign → image is inverted.)

Final Answers: Image distance = +37.5 cm; Height of image = –15 cm (inverted).

Given: $h_0 = 4$ cm, $f = +20$ cm, $u = -15$ cm

Position of Image: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{20} – \frac{1}{15} = \frac{3-4}{60} = -\frac{1}{60}$$ $$v = -60 \, \text{cm}$$ Image is formed 60 cm in front of the lens on the same side as the object.

Size of Image: $$m = \frac{v}{u} = \frac{-60}{-15} = 4 \Rightarrow h_i = 4 \times 4 = 16 \, \text{cm}$$ Nature: Virtual, erect, and magnified (×4).

Final Answer: Position = 60 cm in front of lens; Size = 16 cm; Nature = virtual, erect, magnified.

(i) Principal Focus: It is the point on the principal axis of a convex (diverging) mirror from which parallel rays of light appear to diverge after reflection.

(ii) Focal Length: The focal length (f) is the distance between the pole (P) of the convex mirror and its principal focus (F).

When a ray of light enters a prism, it is refracted at the surface of entry. It then travels through the prism and exits after another refraction. The angle of incidence (∠i) is the angle between the incident ray and the normal at the surface of entry. The angle of deviation (∠D) is the angle between the incident ray and the final emergent ray.

Nature of Mirror: Since the image is real and formed on a screen, the mirror is a concave mirror.

Given: $u = -30$ cm, $v = -60$ cm, $h = 2.4$ cm

Focal Length: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-60} + \frac{1}{-30} = -\frac{3}{60}$$ $$f = -20 \, \text{cm}$$ Height of Image: $$h’ = \frac{v}{u} \times h = \frac{-60}{-30} \times 2.4 = -4.8 \, \text{cm}$$ Nature of Image: The negative sign indicates the image is inverted.

Final Answers: Focal length = –20 cm; Image height = –4.8 cm; Image is inverted.

Given: $f = -12$ cm, $u = -18$ cm, $h = +1.5$ cm

(a) Image Distance: $$\frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{-12} + \frac{1}{18} = \frac{-3+2}{36} = -\frac{1}{36}$$ $$v = -36 \, \text{cm}$$ Distance of image from mirror = 36 cm.

(b) Length of Image: $$h’ = \frac{v}{u} \times h = \frac{-36}{-18} \times 1.5 = 3.0 \, \text{cm (inverted)}$$ Length of image = 3 cm (inverted and magnified).

When object distance is reduced to 10 cm: The object lies between the pole and focus. Reflected rays diverge, forming a virtual, erect, and magnified image behind the mirror. No distinct image will be seen on the screen.

The power of a lens is the ability of the lens to converge or diverge the rays of light passing through it.

It is defined as the reciprocal of the focal length (in metres): $$P = \frac{1}{f}$$ where $P$ is the power (in dioptres, D) and $f$ is the focal length in metres.

Note: If $f$ is positive, the lens is a convex (converging) lens. If $f$ is negative, the lens is a concave (diverging) lens.

(a) From the figure, the light ray bends towards the normal when entering medium B. This happens when light travels from a rarer to a denser medium. Therefore, medium B is optically denser than medium A.

(b) The refractive index of medium B with respect to A: $$\mu_{BA} = \frac{\mu_B}{\mu_A} = \frac{c/v_B}{c/v_A} = \frac{v_A}{v_B}$$ $$\boxed{\mu_{BA} = \frac{v_A}{v_B}}$$

(a) Ray Diagram:



Since the ray passes from a denser medium (diamond) to a rarer medium (water), it bends away from the normal.

(b) Calculation:
Given: $\mu_{\text{diamond}} = 2.42$, $\mu_{\text{water}} = 1.33$ $$\mu_{\text{water/diamond}} = \frac{\mu_{\text{water}}}{\mu_{\text{diamond}}} = \frac{1.33}{2.42} \approx 0.55$$ Refractive index of water w.r.t. diamond = 0.55.

(a) Given: $R = 5$ m, $f = +2.5$ m (convex mirror), $u = -20$ m.

Using mirror formula: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{2}{5} = \frac{1}{v} + \frac{1}{20}$$ $$\frac{1}{v} = \frac{2}{5} – \frac{1}{20} = \frac{8-1}{20} = \frac{9}{20} \Rightarrow v \approx 2.2 \, \text{m}$$ Nature: Virtual and erect (convex mirror always forms virtual, erect, diminished images).
Size: Diminished.



(b) The mirror used by the dentist is a concave mirror. It is used because a concave mirror forms an erect and magnified image of the teeth when the object is placed between the pole and focus, allowing the dentist to examine teeth more clearly.

(a) Convex mirrors are generally used to make search mirrors, as they provide a wider field of view.

(b)


Nature of image formed by a convex mirror:
 • Virtual — cannot be formed on a screen.
 • Erect — upright.
 • Diminished — smaller than the object.

Convex mirrors are ideal for search purposes because they allow a wider area to be observed while producing an erect, diminished image of the hidden objects.

(i) Type of Lens: Positive power (+2D) → convex (converging) lens.

(ii) Focal Length: $$f = \frac{1}{P} = \frac{1}{2} = 0.5 \, \text{m}$$ Image Distance: Using lens formula with $u = -1$ m: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \Rightarrow \frac{1}{0.5} = \frac{1}{v} + 1$$ $$\frac{1}{v} = 2 – 1 = 1 \Rightarrow v = 1 \, \text{m}$$ The image is formed at a distance of 1 m from the lens.

(i) The lens is a convex lens (positive focal length, converging).

(ii) Object Position: $f = 5$ m, $v = +7$ m (real inverted image).
$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \Rightarrow \frac{1}{u} = \frac{1}{7} – \frac{1}{5} = \frac{5-7}{35} = -\frac{2}{35}$$ $$u = -17.5 \, \text{m}$$ The object should be placed 17.5 m to the left of the convex lens.

(iii) Ray Diagram:

(a) Magnification $m = -0.5$ is negative → image is real and inverted. A real inverted image is formed by a converging (convex) lens.





Since $|m| = 0.5 < 1$, the object is placed beyond 2f, forming a real, inverted, and diminished image.

(b) When the laser beam is directed along the principal axis of the convex lens, it passes through the optical centre without bending. A ray passing through the optical centre of any lens continues in the same direction without deviation. That is why the girl observed the beam continuing in the same direction.