Question

A Prism with angle a =60° and produces an angle od deviation of 60°.Then the refractive index of the material of the prism.​

Answer 1

Solution:

Let the refractive index be x , angle(prism) be A and angle of deviation be D.

$\tt\: x = \frac{sin \bigg( \frac{A +D}{2}\bigg)} {sin \bigg( \frac{A}{2} \bigg)}$

$\implies\tt{x = \frac{sin \bigg( \frac{60 + 60}{2} \bigg)} {sin\bigg(\frac{60}{2}\bigg)}}$

$\implies \tt{x = \frac{sin \bigg(\frac{120 \degree}{2} \bigg)}{sin \:30 \degree} }$

$\implies \tt \: x = {\dfrac{sin \: 60 \degree}{sin \: 30 \degree} }$

$\implies\tt\space x={\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}}$

$\implies{\tt{\space x=\sqrt{3}}}$

∴ Refractive index = x =$\sqrt3$

Answer 2

To Find :

The refractive index of the prism .

Given :

• Angle of prism = 60°
• Angle of deviation = 60°

Formula To be used:

$\large \tt{ \bull \: r = \frac{sin( \frac{a+ d}{2} )}{sin( \frac{a}{2} )} }$

where,

• R = refractive index
• a = Angle of prism
• d = Angle of deviation

Solution :

given, Angle of prism = 60°

Angle of deviation = 60°

so, refractive index :

$\tt{ r = \frac{sin( \frac{a + d}{2} )}{sin( \frac{a}{2}) } }$

$\implies \tt{r = \frac{sin( \frac{60 + 60}{2} )}{sin( \frac{60}{2} )}}$

$\implies \tt{r = \frac{sin( \frac{ \cancel{120}}{ \cancel{2}}) }{sin30 \degree} }$

$\tt{ \implies \: r = \frac{sin 60 \degree}{sin30 \degree}}$

$\implies \tt{r = sin60 \degree \times \frac{1}{sin30 \degree} }$

$\tt{ \implies \: r = \frac{ \sqrt{3} }{2} \times 2}$

$\implies \tt{r = \sqrt{3} }$

hence, Refractive index of prism = √3