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# Find lateral shift of light ray while is passes through a parallel glass slab of thickness 10 cm placed in air.The angle of incidence in air is $60^{\circ}$ and angle of refraction in glass is $45^{\circ}$

$(a)\;5 \sqrt 5 \\ (b)\;10 \sqrt 5 \\ (c)\;5 \sqrt 5 \sin 15^{\circ} \\ (d)\;10 \sqrt 2 \sin 15^{\circ}$

$d= \large\frac{t \sin (i-r)}{\cos r}$
$\qquad= \large\frac{10 \sin (60^{\circ}-45^{\circ})}{\cos 45^{\circ}}$
$\qquad= 10 \large\frac{\sin 15}{\cos 45}$
$\qquad = 10 \sqrt 2 \sin 15^{\circ}$
Hence d is the correct answer.

edited Jul 28, 2014