Q30 of 123 Page 412

Light is incident from glass (μ = 1.50) to water (μ = 1.33). Find the range of the angle of deviation for which there are two angles of incidence.


Refractive index of glass is


Refractive index of water is


Critical Angle=


According to the question:


There will be two angles of incidence:


(1) When light ray passes through the normal


Angle of incidence =


Angle of refraction=


Angle of deviation =


(2) When light ray incident at critical angle


(As light ray is passing from glass to water)


⇒


⇒



⇒


⇒ Angle of deviation=


=


=


If angle of incidence increases beyond critical angle, total internal reflection will occur. So, the range of angle of deviation is


More from this chapter

All 123 →
28

Find the maximum angle of refraction when a light ray is refracted from glass (μ = 1.50) to air.

 29

Light is incident from glass (μ = 1.5) to air. Sketch the variation of the angle of deviation δ with the angle of incident i for 0<i<90°.

 31

Light falls from glass (μ = 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.

 32

A point source is placed at a depth h below the surface of water (refractive index = μ).

(a) Show that light escapes through a circular area on the water surface with its center directly above the point source.


(b) Find the angle subtended by a radius of the area on the source.