Question

calculate the angle of minimum deviation for an equilateral prism of refractive index root 3

Answer 1

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

Equilateral prism of refractive index = $\sf \sqrt{3}$

$\checkmark$ To Find:

Calculate the angle of minimum deviation for an equilateral prism of refractive index = $\sf \sqrt{3}$

$\checkmark$ Solution:

We know that,

$\star$ Refractive Index of Prism (μ)

$\blue{\boxed{\sf \mu=\dfrac{sin\left(\dfrac{A+D}{2}\right)}{sin\left(\dfrac{A}{2}\right) } }}$

Here,

μ = Refractive Index

A= Angle of prism

D = Angle of minimum deviation

Given that,

Equilateral prism

Therefore,

A = 60°

According to the Question,

We are asked to find the Angle of minimum deviation

Therefore,

A = 60 °

μ = $\sf \sqrt{3}$

Substituting the above values,

We get,

$\implies \sf \sqrt{3} =\dfrac{sin\left(\dfrac{60+D}{2}\right)}{sin\left(\dfrac{60}{2}\right)}$

$\implies \sf \sqrt{3} =\dfrac{sin\left(\dfrac{60+D}{2}\right)}{sin(30^{\circ})}$

$\rm We \ know \ that, \\\\ \longrightarrow sin 30 ^{\circ}= \dfrac{1}{2}$

$\implies \sf \sqrt{3} =\dfrac{sin\left(\dfrac{60+D}{2}\right)}{\dfrac{1}{2}}$

$\sf \implies \dfrac{\sqrt{3} }{2}=sin\left( \dfrac {60+D}{2}\right)$

Taking sine function to the other side it becomes sine inverse

Therefore,

$\sf \implies sin^{-1} \left(\dfrac{\sqrt{3} }{2}\right) =\left( \dfrac {60+D}{2}\right)$

Taking the 2 of denominator of the right side fraction to the other side it becomes numerator of the left side fraction

Hence,

$\sf \implies 2sin^{-1} \left(\dfrac{\sqrt{3} }{2}\right) =60+D$

$\rm We \ know \ that, \\\\ \longrightarrow sin^{-1} \left(\dfrac{\sqrt{3} }{2} \right )=60^{\circ} \\\\ Therefore, \\\\ 2sin^{-1} \left(\dfrac{\sqrt{3} }{2} \right )= 2 \times 60^{\circ} \\\\ Hence, \\\\ 2sin^{-1} \left(\dfrac{\sqrt{3} }{2} \right )=120^{\circ}$

Therefore,

$\sf \implies 120^{\circ}=60^{\circ}+D$

$\sf \implies 60^{\circ}+D=120^{\circ}$

$\sf \implies D=120^{\circ}-60^{\circ}$

Hence,

$\sf \implies D=60^{\circ}$

Therefore,

The angle of minimum deviation = 60 °