A beam of light falls on a glass plate $(μ = 3 / 2)$ of thickness 6.0 cm at an angle of $60^{0}$ . Find the lateral shift of the beam on passing through the plate (in cm) ( Take $\sqrt{3} = 1.7 a n d \sqrt{2} = 1.4$ )

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answer is 3.

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Detailed Solution

$\cos r = \frac{t}{y} \Rightarrow y = \frac{t}{\cos r} a n d \sin (i - r) = \frac{x}{y}$

$\Rightarrow x = \frac{t}{\cos r} [\sin i \cos r - \cos i \sin r]$

$x = t [\sin i - \cos i \tan r],$

$μ = \frac{\sin i}{\sin r} \Rightarrow \sin r = \frac{\sin 60^{\circ}}{μ}$

$\sin r = \frac{1}{\sqrt{3}}; \cos r = \frac{\sqrt{2}}{\sqrt{3}}; \tan r = \frac{1}{\sqrt{2}}$

$∴ x = 6 [\sin 60^{\circ} - \cos 60^{\circ} \frac{1}{\sqrt{2}}]$

$= 6 [\frac{\sqrt{3}}{2} - \frac{1}{2} \frac{1}{\sqrt{2}}] = 3 [1.7 - 0.7] = 3 c m$

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