A ray is incident at an angle of incidence i on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ. then the angle of incidence is nearly equal to:

Question

A ray is incident at an angle of incidence i on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ. then the angle of incidence is nearly equal to: (A)  (A/2μ) (B)  (2A/μ) (C) μ A (D)  (μ A/2)

Tardigrade Admin · Accepted Answer

Light ray emerges from another surface normally, hence e(angle of emergence) = 0

\Rightarrow r_{2} = 0

r_{1} + r_{2} = A

\Rightarrow r_{1} = A

Applying Snell’s law on first surface,

1. s in i = μ s in r_{1}

For small angle ,

s in θ \approx θ

Hence,

i = μ A

Q. A ray is incident at an angle of incidence $i$ on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $μ$ . then the angle of incidence is nearly equal to:

$\frac{A}{2 μ}$

$\frac{2 A}{μ}$

$μ A$

$\frac{μ A}{2}$

Light ray emerges from another surface normally, hence e(angle of emergence) = 0

$\Rightarrow r_{2} = 0$
$r_{1} + r_{2} = A$
$\Rightarrow r_{1} = A$
Applying Snell’s law on first surface,
$1. s in i = μ s in r_{1}$
For small angle , $s in θ \approx θ$ Hence,
$i = μ A$

Q. A ray is incident at an angle of incidence i on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ. then the angle of incidence is nearly equal to:

2μA​

μ2A​

μA