Question

A light ray is incident on air - liquid interface at 45° and is refracted at 30°.What is the refractive index of the liquid?
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Answer 1

Answer:

Given:

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

$1.41$

Explanation:

$we \: \: know \: \: \boxed{Snell's \: law \: \: = > \: \: \mu = \frac{sin \: i}{sin \: r} } \\ Given, \: \: \: i = 45° \: \: \: and \: \: \: r = 30° \\ Now, \: \: \: the \: \: refractive \: \: index \: \: of \: \: the \: \:liquid \\ \: \: \: \: \: \mu = \frac{sin45°}{sin30°} \\ \: = \frac{ \frac{1}{ \sqrt{2} } }{ \frac{1}{2} } \\ \: \: \: \: \: \: \: \: \: = \frac{1}{ \sqrt{2} } \times 2 \\ = \sqrt{2} \\ \: \: \:\: \:= \boxed{1.41}$