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Prism formula | Derivation of prism formula

Prism formula | Derivation of prism formula :- As we have seen in article Condition for minimum deviation in a prism, in case of minimum deviation,

i₁ = i₂ = i (let) and

r₁ = r₂ = r (let)

Now as $\displaystyle \delta ={{i}_{1}}+{{i}_{2}}-A$ , so in case of minimum deviation,

$\displaystyle {{\delta }_{\min }}=i+i-A=2i-A$

$\displaystyle \Rightarrow i=\frac{A+{{\delta }_{\min }}}{2}$ …..(1)

Also $\displaystyle A={{r}_{1}}+{{r}_{2}}$ , in case of minimum deviation,

$\displaystyle A=r+r=2r$

$\displaystyle \Rightarrow r=\frac{A}{2}$ …..(2)

Using Snell’s Law,

$\displaystyle \mu =\frac{\sin i}{\sin r}=\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$

$\displaystyle \Rightarrow \mu =\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$ …..(3)

This is called prism formula. It is used for accurate determination of refractive index of a transparent medium, of which the prism is made.

Illustration 1.

A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. The angle of the prism is 60° . Find the angle of minimum deviation and refractive index of the material of the prism.

Solution:-

Here i₁ = 50°, A = 60°, δ_m = ? and μ = ?

In case of minimum deviation, i₁ = i₂ = i = 50°

Using δ_m= 2i – A

⇒ δ_m= 2(50°) – 60° = 40°

Now using $\displaystyle \mu =\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$ , we get

$\displaystyle \mu =\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}=\frac{\sin \left( \frac{{{60}^{\circ }}+{{40}^{\circ }}}{2} \right)}{\sin \left( \frac{{{60}^{\circ }}}{2} \right)}=\frac{\sin {{50}^{\circ }}}{\sin {{30}^{\circ }}}=\frac{0.7660}{0.5000}$

$\displaystyle \Rightarrow \mu =1.532$

Illustration 2.

For a glass prism $μ = \sqrt{3}$ the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Solution:-

Here it is given that δ_min = A

Using $\displaystyle \mu =\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$ , and putting δ_min = A, we get

$\displaystyle \sqrt{3}=\frac{\sin \left( \frac{A+A}{2} \right)}{\sin \left( \frac{A}{2} \right)}=\frac{\sin A}{\sin \left( \frac{A}{2} \right)}=\frac{2\sin \left( \frac{A}{2} \right)\cos \left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)}$