An isosceles prism of angle $120^{\circ}$ has a refractive index $\sqrt{2}$ . Two parallel monochromatic rays enter the prism parallel to each other in air as shown. in figure. Find the angle between the emerging rays

30^{\circ}

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

45^{\circ}

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

60^{\circ}

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

90^{\circ}

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

Open in App

Solution

The correct option is A $30^{\circ}$

$\frac{s i n r}{s i n 30^{\circ}} = \sqrt{2} s i n 30^{\circ} s i n r = \frac{1}{\sqrt{2}} \Rightarrow r = 45^{\circ}$
Angle $∠ = 45^{\circ} - 30^{\circ} = 15^{\circ}$
Required angle = $2 d = 30^{\circ}$

Suggest Corrections

Similar questions

An isosceles prism of angle $120^{\circ}$ has a refractive index 1.44. Two parallel monochromatic rays enter the prism parallel to each other in air as shown in the figure. The rays emerging from the opposite faces.

Q. An isosceles prism of angle

120^{o}

has a refractive index 1.44. Two parallel monochromatic rays enter the prism parallel to each other in air as shown in Fig 13.76. The rays emerging from the opposite faces :

Q. Two parallel light rays pass through an isosceles prism of refractive index

\sqrt{\frac{3}{2}}

as shown in the figure. The angle between the two emergent rays is :

Q. The prism shown in the figure has a refractive index of

1.60

and angle

A

30^{\circ}

. Two parallel light rays

P

and

Q

enter the prism. The angle between the light rays, when they emerge is:

[

Given:

{sin}^{- 1} (0.8) = 53^{\circ}]

Two parallel light rays are incident at one surface of a prism of refractive index 1.5 as shown in figure. The angle between the emergent rays is nearly

Join BYJU'S Learning Program

An isosceles prism of angle 120∘ has a refractive index √2. Two parallel monochromatic rays enter the prism parallel to each other in air as shown. in figure. Find the angle between the emerging rays

The correct option is A 30∘ sin rsin30∘=√2sin30∘sin r=1√2⇒r=45∘ Angle ∠=45∘−30∘=15∘ Required angle =2d=30∘

An isosceles prism of angle $120^{\circ}$ has a refractive index $\sqrt{2}$ . Two parallel monochromatic rays enter the prism parallel to each other in air as shown. in figure. Find the angle between the emerging rays

The correct option is A $30^{\circ}$

$\frac{s i n r}{s i n 30^{\circ}} = \sqrt{2} s i n 30^{\circ} s i n r = \frac{1}{\sqrt{2}} \Rightarrow r = 45^{\circ}$
Angle $∠ = 45^{\circ} - 30^{\circ} = 15^{\circ}$
Required angle = $2 d = 30^{\circ}$