Question

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.

Answer 1

Deviation angle range is $\textrm{0 to 37.27\°}$

Explanation:

Given that,

Refractive angle of glass $\mu_g = 1.5$ $= \frac{3}{2}$

Refractive angle of water $\mu_w = 1.33 = \frac{4}{3}$

According to the question for two angle incidence

(a) When light ray travels straight along or parallel to the normal then,

Incidence angle $i = 0$

Refraction angle $i_r = 0$

Deviation angle $i_d = 0$

(b) When light ray is incident at critical angle( at this angle refraction angle becomes 90˚) and travels through glass and then water,

According to snell's law,

$frac{\sin i_c}{\sin i_r} = \frac{\mu_w}{mu_g}$  

$\frac{\sin i_c}{\sin i_r} = \dfrac{\frac{4}{3}}{\frac{3}{2}}$

$\sin i_c = \frac{8}{9}$

$i_c = \sin^{-1}(\frac{8}{9}) = 67.73\°$

Deviation angle $= 90\° - 67.73\° = 37.27\°$

So the deviation angle range is $\textrm{0 to 37.27\°}$

Answer 2

Range of Deviation is 0 - 37.27°

Explanation:

$\mu_g = 1.5 = \frac {3}{2}\ and\ \mu_w = 1.33 = \frac {4}{3}$

For the two angle of incidence,

1. When light passes straight through normal,

Angle of incidence = 0°, Angle of refraction = 0°, Angle of Deviation = 0°  

2. When light is incident at critical angle,

$\frac {sinC}{sin r} = \frac {\mu_w}{\mu_g}$

C = $sin^-^1(\frac {8}{9})$

= 62.73°

So, Angle of deviation = 90° - C

= $90 - sin^-^1(\frac {8}{9})$

= $cos^-^1(\frac {8}{9})$

= 37.27°

Here, if the angle of incidence is increased beyond critical angle, total internal reflection occurs and deviation decreases. So, the range of deviation is 0 to $cos^-^1(\frac {8}{9}).$