A ray of light enters a glass slab of thickness 10 cm and refractive index 1.5 at an angle of 45 degrees. Calculate the lateral shift of the ray.

B. 3.2 cm — The lateral shift is given by the formula s = t (i - r) / (r). First, find the angle of refraction r using Snell's law.

JEE Main · PhysicsHard

$A ray of light enters a glass slab of thickness $t = 10 \text{ cm}$ and refractive index $\mu = 1.5$ at an angle of incidence $i = 45^\circ$. Calculate the lateral shift $s$.$

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Correct answer: B — $3.2 \text{ cm}$

Using

sin (4 5^{\circ}) = 1.5 sin (r)

, we get

sin (r) = 1/ (1.5 2) \approx 0.471

. Then

r \approx 28. 1^{\circ}

. Lateral shift

s = 10 sin (4 5^{\circ} - 28. 1^{\circ}) / cos (28. 1^{\circ}) \approx 3.2 cm