Question

Light propagating in air strikes the surface of a lake. The reflected ray makes 62° angle with the normal. What angle does the refracted ray in the water makes with the normal. (Hint: index of refraction for water 1.33) (a) 420 (b) 48° (c) 540 (d) 58° me= 9.11 × 10-31 kg 80-8.85 x 10-12 C/N.m2 h= 6.63 × 10-34 J-s m,-1.67 x 10-27 kg k-8.988 10 N -m/C e= 1.6× 10-19c o4t x 10-7 N/A Pa 1.69 × 10-4 Ω-m.

Answer 1

Answer:

(a) 42°

Explanation:

According to the law of reflection:

Angle of incidence = Angle of reflection

So, angle of reflection = 62°

Angle of incidence = 62°

For refraction,

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,  

${\theta_i}$  is the angle of incidence  ( 62.0° )

${\theta_r}$ is the angle of refraction  ( ? )

${n_r}$ is the refractive index of the refraction medium  (water, n=1.33)

${n_i}$ is the refractive index of the incidence medium (air, n=1)

Hence,  

$1\times {sin62.0^0}={1.33}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.6639$ = 42°