Question

A prism with an angle A = 60degree produces an angle of minimum deviation D of 30degree Find the refractive index of the material of the prism​

Answer 1

$GIVEN:-$

• $A = 60^{\circ}$
• $D = 30^{\circ}$

$\\ TO\:FIND:-$

• Refractive index of the material of the prism

$\\ SOLUTION:-$

★ Refractive index

$\implies \:\: n = \dfrac{ sin\bigg(\dfrac{A+D}{2}\bigg) }{ sin\bigg(\dfrac{A}{2}\bigg) }$

$\implies \:\: n = \dfrac{ sin\bigg(\dfrac{60+30}{2}\bigg) }{ sin\bigg(\dfrac{60}{2}\bigg) }$

$\implies \:\: \dfrac{ sin\bigg(\dfrac{90}{2}\bigg) }{ sin\:30 }$

$\implies \:\: \dfrac{ sin\:45^{\circ}}{sin\:30^{\circ}}$

$\implies \:\: \dfrac{ \dfrac{1}{\sqrt{2}} }{\dfrac{1}{2}}$

$\implies \:\: \dfrac{ 1}{\sqrt{2}} \times \dfrac{2}{1}$

$\implies \:\:{\boxed{ \sqrt{2}}}$

$\therefore$ The refractive index of the material of the prism $\sqrt{2}$ = $1.414$