Question

If the polarizing angle of a piece of glass for green light is
54.74°, then the angle of minimum deviation for an
equilateral prism made of same glass is
$\sf (Given, \tan \: 54.74^{ \circ} = 1.414)$

​

Answer 1

Answer:

δ=90°−60° =30° is the answer

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Explanation:

Answer 2

Given :

Polarizing angle of a glass = 54.74°

To Find :

Angle of minimum deviation for an equilateral prism made of same glass.

Solution :

❒ In this question, both piece of glass and prism are made of same material means both have same refractive index.

Let's find refractive index of glass first.

As per Brewster's law :

• n = tan $\sf{i_p}$

where, n denotes refractive index and $\sf{i_p}$ denotes polarizing angle.

➙ n = tan (54.74°)

➙ n = 1.414

➙ n = √2

★ The refractive index of the material of the prism is given by

$:\implies\sf\:n=\dfrac{\bigg[\dfrac{(A+\delta_m)}{2}\bigg]}{sin\bigg(\dfrac{A}{2}\bigg)}$

• n denotes refractive index
• A denotes angle of prism
• $\sf{\delta_m}$ denotes angle of minimum deviation

For an equilateral prism, A = 60°

By substituting the values;

$:\implies\sf\:\sqrt{2}=\dfrac{sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]}{sin\bigg(\dfrac{60^{\circ}}{2}\bigg)}$

$:\implies\sf\:\sqrt{2}=\dfrac{sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]}{sin\:30^{\circ}}$

$\sf:\implies\:\sqrt{2}\times sin\:30^{\circ}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:\sqrt{2}\times\dfrac{1}{2}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:\dfrac{1}{\sqrt2}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

• sin 45° = 1/√2

$:\implies\sf\:sin\:45^{\circ}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:45^{\circ}=\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:60^{\circ}+\delta_m=90^{\circ}$

$:\implies\:\underline{\boxed{\bf{\purple{\delta_m=30^{\circ}}}}}$