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 slab. The perpendicular distance between the incident ray and the emergent ray is

5 / \sqrt{3} c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{3} / 5 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 / 3 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 / 2 c m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $5 / 2 c m$
Using snell's law:

$\frac{s i n 60^{\circ}}{s i n r_{1}} = \sqrt{3} \Rightarrow s i n r_{1} = \frac{s i n 60^{\circ}}{\sqrt{3}} \Rightarrow s i n r_{1} = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} = \frac{1}{2} \Rightarrow r_{1} = 30^{\circ} Now, we know that, s i n (i_{1} - r_{1}) = \frac{d}{5} \Rightarrow d = 5 s i n (i_{1} - r_{1}) \Rightarrow d = 5 s i n (60^{\circ} - 30^{\circ}) \Rightarrow d = 5 s i n 30^{\circ} = \frac{5}{2} c m$

Suggest Corrections

Similar questions

Q. A ray of light enters a rectangular glass slab of refractive index

\sqrt{3}

at an angle of incidence

60^{O}

. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

Q. A ray of light is incident on a glass slab. The light ray travels distance of 5 cm inside the glass slab before emerging out of the slab. If the incident ray suffers a deviation of

30^{\circ}

after the first referaction, the perpendicular distance between incident and the emergent ray is :

Q. Refractive index of a rectangular glass slab is

μ = \sqrt{3}

. A light ray incident at an angle

60^{\circ}

is displaced laterally through

2.5 cm

. Distance travelled by light in the slab is

A ray of light is incident on a glass-slab from air at an angle . Thickness and refractive index of glass-slab are‘t’ and ‘ $μ_{a}$ ’ respectively, Find the lateral displacement of the emergent ray respective to incident ray.

Q. A ray of light is incident on a glass plate. The light ray travels distance of 5 cm inside the glass plate before emerging out of the glass plate. If the incident ray suffers a deviation of

30^{0}

. the perpendicular distance between incident and the emergent ray is

Join BYJU'S Learning Program

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 slab. The perpendicular distance between the incident ray and the emergent ray is

The correct option is D 5/2cmUsing snell's law: sin 60∘sin r1=√3⇒ sin r1=sin 60∘√3⇒sin r1=√32×1√3=12⇒r1=30∘Now, we know that,sin(i1−r1)=d5⇒d=5 sin(i1−r1)⇒d=5sin(60∘−30∘)⇒d=5sin30∘=52cm

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 slab. The perpendicular distance between the incident ray and the emergent ray is