Question

A ray of light is incident at an angle of 45°C on one face of a rectangular glass slab of thickness 10cm and refractive index 1.5. Calculate lateral shift.

Answer 1

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your answer is in attachment ...✌ ✌

Answer 2

Answer: The lateral  shift is 0.03 m.

Explanation:

Given that,

Incident angle = 45°

Thickness t = 10 cm

Refractive index n = 1.5

We know that,

$n =\dfrac{sin\ i}{sin\ r}$

$1.5 = \dfrac{sin\ 45^{\circ}}{sin\ r}$

$sin\ r= \dfrac{1}{\sqrt{2}\times1.5}$

$sin\ r= 0.471$

$r = sin^{-1}( 0.471)$

$r = 28^{\circ}$

The lateral shift is

$L_{s}= \dfrac{t\ sin(i-r)}{\cos r}$

$L_{s}= \dfrac{0.1\times\ sin(45^{\circ}-28^{\circ})}{\cos 28^{\circ}}$

$L_{s}=0.03\ m$

Hence, The lateral  shift is 0.03 m.