A ray of light enters a rectangular glass slab of refractive index 3 at an angle of incidence 60∘. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and emergent rays is

sin⁡60∘sin⁡r1=3⇒sin⁡r1=sin⁡60∘3=32×13=12sin⁡i1−r1=d5⇒d=5sin⁡i1−r1 ⇒d=5 sin⁡60∘− 30∘=5 sin⁡30∘=52cm

A ray of light enters a rectangular glass slab of refractive index 3 at an angle of incidence 60∘. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and emergent rays is

sin⁡60∘sin⁡r1=3⇒sin⁡r1=sin⁡60∘3=32×13=12sin⁡i1−r1=d5⇒d=5sin⁡i1−r1 ⇒d=5 sin⁡60∘− 30∘=5 sin⁡30∘=52cm

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A ray of light enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{\circ}$ . It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and emergent rays is

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$5 \sqrt{3} cm$

$\frac{5}{2} cm$

$5 \sqrt{3 / 2} cm$

5 cm

answer is B.

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Detailed Solution

$\frac{\sin 60^{\circ}}{\sin r_{1}} = \sqrt{3}$

$\Rightarrow \sin r_{1} = \frac{\sin 60^{\circ}}{\sqrt{3}} = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} = \frac{1}{2}$

$\sin (i_{1} - r_{1}) = \frac{d}{5}$

$\Rightarrow d = 5 \sin (i_{1} - r_{1}) \Rightarrow d = 5 \sin (60^{\circ} - 30^{\circ}) = 5 \sin 30^{\circ} = \frac{5}{2} cm$

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A ray of light enters a rectangular glass slab of refractive index 3 at an angle of incidence 60∘. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and emergent rays is

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A ray of light enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{\circ}$ . It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and emergent rays is