Ray Optics Question 4
Question 4 - 25 January - Shift 1
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)
Solution:
Formula: Laws of Refraction
$\sin c=\frac{1}{\sqrt{2}}$
$c=45^{\circ}$
$\sin c=\mu \sin \theta$
$\frac{1}{\sqrt{2}}=\sqrt{2} \sin \theta$
$\theta=30^{\circ}$
Lateral displacement:
$x=t \sin (i-r) \sec r$
$x=\sqrt{3} \sin (45^{\circ}-30^{\circ}) \sec 30^{\circ}$
$x=\sqrt{3}(0.26)(\frac{2}{\sqrt{3}})$
