Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

$120^{\circ}$
$60^{\circ}$
$45^{\circ}$
$90^{\circ}$
$30^{\circ}$

Question

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

$120^{\circ}$
$60^{\circ}$
$45^{\circ}$
$90^{\circ}$
$30^{\circ}$

Toppr Admin · Accepted Answer

Angle of incident at the boundary - 60By snells law-x21D2-x221A-3sin60-x2218-1-5sinr-x21D2-r-90-x2218-Thus the angle of deviation as seen in the figure is given as 90o-x2212-60o-30o

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :120∘60∘45∘90∘30∘

Angle of incident at the boundary = 60By snells law⇒√3sin60∘=1.5sinr⇒r=90∘Thus the angle of deviation as seen in the figure is given as 90o−60o=30o

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

$120^{\circ}$
$60^{\circ}$
$45^{\circ}$
$90^{\circ}$
$30^{\circ}$

Angle of incident at the boundary = 60
By snells law
$\Rightarrow \sqrt{3} sin 60^{\circ} = 1.5 sin r$
$\Rightarrow r = 90^{\circ}$
Thus the angle of deviation as seen in the figure is given as $90^{o} - 60^{o} = 30^{o}$