Question 95 of 295

A ray of light is incident from air on a glass plate having thickness $\sqrt{3}cm$ and refractive index $\sqrt{2}$. 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 {{10}^{-}}^{2}cm . (given $\sin 15^{\circ}=0.26$)

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Answer: 52

Let $C$ be the critical angle.
Then, $sinC=\dfrac{1}{\mu }$, where, $\mu$ is the refractive index of glass.
So, $sinC=\dfrac{1}{\sqrt{2}}\Rightarrow$$C=45^{\circ}$
As the angle of incidence is equal to the critical angle, so using Snell's law,
$sinC=\mu \sin \theta$
$\Rightarrow \dfrac{1}{\sqrt{2}}=\sqrt{2}\sin \theta$
$\Rightarrow \theta =30^{\circ}$
Now, lateral displacement of the ray is
$x=t\sin (i-r)\sec r$
$\Rightarrow x=\sqrt{3}\sin \left(45^{\circ}-30^{\circ}\right)\sec 30^{\circ}$
$\Rightarrow x=\sqrt{3}(0.26)\left(\dfrac{2}{\sqrt{3}}\right)$
$\Rightarrow x=0.52cm$
$\Rightarrow x=52\times {10}^{-2}cm$