Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is :

Question

Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is : (A) 1.732 (B) 1.333 (C) 1.414 (D) 2.732

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/81fcdd65dfb0a6b9483ff95ccd5ec701-.png" /> i =45° D = i - r 15°=45- r ⇒ r =30° n 1 sin i = n 2 textsin r 1 sin 45°=μ sin 30° (1/√2)=μ (1/2) μ=√2=1.414

Q. Light enters from air into a given medium at an angle of $4 5^{\circ}$ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of $1 5^{\circ}$ from its original direction. The refractive index of the medium is :

$1.732$

$1.333$

$1.414$

$2.732$

$i = 4 5^{\circ}$
$D = i - r$
$1 5^{\circ} = 45 - r \Rightarrow r = 3 0^{\circ}$
$n_{1} sin i = n_{2} sin r$
$1 sin 4 5^{\circ} = μ sin 3 0^{\circ}$
$\frac{1}{2} = μ \frac{1}{2}$
$μ = 2 = 1.414$

Q. Light enters from air into a given medium at an angle of 45∘ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15∘ from its original direction. The refractive index of the medium is :

1.732

1.333

1.414

2.732