Question

A narrow beam of light is incident normally an one face of an equilateral prism and finally emerges from the prism. Thr prism is now surrounded by water. What is the angle between the directions of the emergent beam in the two cases?(Mg=1.45, Mw=1.33)

Answer 1

angle of deviation in prism is:S = i + e – Alight incident normally on the face .i.e.i = 0 and  r1 = 0 r2 = 60 degree.r2 > critical angle .i.e = arcsin(1/1.5) = 41.81 degree. S = 0 – 1 + 60 = degree.

Answer 2

Concept:

When any light ray goes from an optically denser medium to an optically rarer medium and is incidence at a greater angle than the critical angle for the two media, the ray completely reflects back into the medium, according to reflection laws. This is referred to as complete inward reflection.

Given:

Refractive index of glass Mg = 1.45

Refractive index of water Mw = 1.45

Incidence angle ( i ) = 90°

Find:

Difference between the angle of the emergent rays in the two given cases.

Solution:

When the prism is in air

Incidence angle ( i ) = 90°

Mg = 1.45

Mw = 1.33

For critical angle

Mg / Ma = 1 / Sin C

Ma is refractive index of air (surrounding) which is 1

Mg = 1 / Sin C

Sin C = 1 / Mg

Sin C = 1 / 1.45

Sin C = 0.6896

C = Sin⁻¹(0.6896)

C = 43.60°

As the light ray pass from glass to air angle of incidence is 60° and critical angle also is 60°, So here i > C hence, Total Internal reflection will take place.

When prism is surrounded by water

Incidence angle ( i )= 60°

For critical angle

Mg / Mw = 1 / Sin C

Sin C = Mw / Mg

Sin C = 1.33 / 1.45

Sin C = 0.917

C = Sin⁻¹(0.917)

C = 66.5°

here incidence angle is less than critical angle so refraction will take place.

Using snells law

Mg / Mw = Sin r / Sin i

1.45 / 1.33 = Sin r / Sin 60°

1.09 = Sin r / (√3 / 2)

Sin r = 0.94

r = Sin⁻¹(0.94)

∴ r = 71°

Thats why angle between two emergent ray is = 180° - 60° - 71°

angle between two emergent ray is = 49°

Hence, the difference between the angle of the emergent rays in the two given cases is 49°.

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