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ICSE Class 10 Physics Question 13 of 25

Model Paper 1 — Question 2

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Question

Question 4(ii)

The adjacent diagram represents a glass slab of refractive index 1.5. If PA, PB, PC and PD represent the rays of light from a point P at the bottom of the block, draw the approximate directions of these rays as they emerge out of the glass slab. (sin 42° = 2/3)

The adjacent diagram represents a glass slab of refractive index 1.5. If PA, PB, PC and PD represent the rays of light from a point P at the bottom of the block, draw the approximate directions of these rays as they emerge out of the glass slab. Concise Physics Solutions ICSE Class 10.
Answer

Given,

Refractive index of glass with respect to air = aμg = 1.5 = 1gμa\dfrac{1}{_\text g\text μ_\text a}

Angle made by PA with normal = 0°

Angle made by PB with normal = 30°

Angle made by PC with normal = 42°

Angle made by PD with normal = 60°

sin 42° = 2/3

The adjacent diagram represents a glass slab of refractive index 1.5. If PA, PB, PC and PD represent the rays of light from a point P at the bottom of the block, draw the approximate directions of these rays as they emerge out of the glass slab. Concise Physics Solutions ICSE Class 10.

(i) For ray PA

i = 0°

sin r=sin igμa=sin 0°×1.5=0\text {sin r} = \dfrac{\text{sin i}}{{_g\text{μ}_a}}=\text {sin }0°\times 1.5 =0

⟹ r = sin-1(0) = 0°

Ray PA will not deviate and emerge straight.

(ii) For PB

i = 30°

sin r=sin igμa=sin 30°×1.5=0.5×1.5=0.75\text {sin r} = \dfrac{\text{sin i}}{{_\text{g}\text{μ}_a}}=\text {sin}\ 30°\times 1.5 = 0.5\times1.5=0.75

⟹ r = sin-1(0.75) ≈ 49°

Ray PB will emerge at an angle of 49° with the normal.

(iii) For PC

i = 42°

sin r=sin igμa=sin 42°×1.5=23×1.5=1\text {sin r} = \dfrac{\text{sin i}}{{_\text{g}\text{μ}_a}}=\text {sin}\ 42°\times 1.5 = \dfrac{2}{3} \times1.5=1

⟹ r = sin-1(1) = 90°

This is the condition for critical angle.

Ray PC will graze along the surface.

(iv) For PD

i = 60°

sin r=sin igμa=sin 60°×1.5=32×1.5≈1.3\text {sin r} = \dfrac{\text{sin i}}{{_\text {g}\text{μ}_a}}=\text {sin}\ 60°\times 1.5 = \dfrac{\sqrt 3}{2} \times1.5≈1.3

Here,

sin r > 1 which a condition of total internal reflection.

Ray PD will suffer total internal reflection.