A light ray is incident at an angle of 45- with the normal to a -2 cm thick plate -2-0- Find the shift in the path of the light as it emerges out from the plate-

v_1 sin 45^∘ = v_2 sin r or, sin r=1/2√(2) or, cos r=√(7)/2√(2) or, d=t/cos r[sin(j r)sin r] =1/2√(2) =√(2)× 2√(2)/√(7) [ 1 ×√(7)/√(2)× 2√(2) 1/√(2)×1/2√(2) ] ...

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Byju's Answer

Standard XII

Physics

Lateral Displacement

A light ray i...

Question

A light ray is incident at an angle of $45^{\circ}$ with the normal to a $\sqrt{2} cm$ thick plate $(μ = 2.0)$ . Find the shift in the path of the light as it emerges out from the plate.

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Solution

$v_{1} s i n 45^{\circ} = v_{2} s i n r$

or, $s i n r = \frac{1}{2 \sqrt{2}}$

or, $c o s r = \frac{\sqrt{7}}{2 \sqrt{2}}$

or, $d = \frac{t}{c o s r} [s i n (j - r) s i n r]$

$= \frac{1}{2 \sqrt{2}}$

$= \frac{\sqrt{2} \times 2 \sqrt{2}}{\sqrt{7}}$

$[\frac{1 \times \sqrt{7}}{\sqrt{2} \times 2 \sqrt{2}} - \frac{1}{\sqrt{2}} \times \frac{1}{2 \sqrt{2}}]$

$= \frac{4}{\sqrt{7} \times 4} [\sqrt{7} - 1]$

$= \frac{\sqrt{7} - 1}{\sqrt{7}}$

= 0.63 cm

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A light ray is incident at an angle of 45∘ with the normal to a √2 cm thick plate (μ=2.0). Find the shift in the path of the light as it emerges out from the plate.

v1 sin 45∘=v2 sin r or, sin r=12√2 or, cos r=√72√2 or, d=tcos r[sin(j−r)sin r] =12√2 =√2×2√2√7 [1×√7√2×2√2−1√2×12√2] =4√7×4[√7−1] =√7−1√7 = 0.63 cm

A light ray is incident at an angle of $45^{\circ}$ with the normal to a $\sqrt{2} cm$ thick plate $(μ = 2.0)$ . Find the shift in the path of the light as it emerges out from the plate.