JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Geometrical Optics

A ray of light is incident from air on a glass plate having thickness

3 cm

and refractive index

2

. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is

____

\times 1 0^{-}^{2} cm

. (given

sin 1 5^{\circ} = 0.26

)

Answer: 52

Solution

At the glass-air boundary the critical angle satisfies

sin c = \frac{1}{μ} = \frac{1}{2}

, so

c = 4 5^{\circ}

, which is the angle of incidence on the plate. Snell's law at entry,

sin 4 5^{\circ} = 2 sin r

, gives

sin r = \frac{1}{2}

and

r = 3 0^{\circ}

. The lateral shift through a plate of thickness

t

d = \frac{t sin ( i - r )}{cos r} = \frac{3 sin 1 5 ^{\circ}}{cos 3 0 ^{\circ}} = \frac{3 \times 0.26}{3 /2} = 0.52 cm

. Written as

x \times 1 0^{- 2} cm

, this gives

x = 52

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Geometrical Optics

Solution

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