Question

A brainiest can do this question.but dont give worst answers and get free points.so who has talent please solve. a light ray is incident on air liquid interface at 45° and is refracted at 30° what is the refractive index of the liquid ? for what angle of indience will the angle between reflected ray and refracted ray be 90°.​

Answer 1

Angle of incidence i = 45°

Angle of refraction r = 30°

 

refractive index =n=sin i /sin r 

=sin45°/sin 30°

=1/√2/1/2⇒2/√2⇒√2=1.4.14

Given the angle between the reflected and

refracted ray = 90°

 It means angle of reflection (r) + angle of incidence (i) = 90°

 Angle of refraction (r) = (90° - i) 

But refractive index n =  sin i/ sin r

=sin i/ sin(90-i)

=sin i/ cos i (∵sin(90-θ)=cos θ)  

From Natural tangent tables tan 54.7° = 1.414 = tan i

The angle of incidence i = 54.7°

Answer 2

Angle of incidence i = 45°

Angle of refraction r = 30°

refractive index =n=sin i /sin r

=sin45°/sin 30°

=1/√2/1/2⇒2/√2⇒√2=1.4.14

Given the angle between the reflected and

refracted ray = 90°

It means angle of reflection (r) + angle of incidence (i) = 90°

Angle of refraction (r) = (90° - i)

But refractive index n = sin i/ sin r

=sin i/ sin(90-i)

=sin i/ cos i (∵sin(90-θ)=cos θ)

From Natural tangent tables tan 54.7° = 1.414 = tan i

The angle of incidence i = 54.7°