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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.

Example 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

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

$\displaystyle \Rightarrow \mu =1.532$

Example 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

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

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

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

$\displaystyle \Rightarrow \frac{A}{2}={{30}^{\circ }}$

$\displaystyle \Rightarrow A={{60}^{\circ }}$

Example 3.

A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is 𝛿 = 60° (see Figure-1). The angle of minimum deviation for red light from the same prism is 𝛿_min = 30° (see Figure-2). The refractive index of the prism material for blue light is √3. Which of the following statement(s) is (are) correct ?

(A) The blue light is polarized in the plane of incidence.

(B) The angle of the prism is 45°.

(D) The angle of refraction for blue light in air at the exit plane of the prism is 60°.

[JEE Advanced 2023 Paper 1 Online]

Solution :-

For blue light, they say “no reflection occurs,” which implies that the angle of incidence is Brewster’s angle — because at Brewster’s angle, all light is transmitted, especially when the incident light is already plane polarized.

Brewster’s Law says:

where is the Brewster’s angle and μ is the refractive index of the denser medium with respect to air.

Brewster’s law tells us that when the angle of incidence is equal to Brewster’s angle, the reflected and refracted rays are perpendicular, and reflected light is fully polarized.

🌟 When incident light is already plane polarized, and its polarization direction is parallel to the plane of incidence, and it hits the surface at Brewster’s angle, then the reflected ray vanishes completely. This is why there is no reflected ray in figure-1.

Now for blue light :

$\displaystyle {{\mu }_{blue}}=tan{{i}_{p}}$

$\displaystyle \Rightarrow \sqrt{3}=tan{{i}_{p}}$

Hence blue light is incident at an angle of incidence i₁ =

Applying Snell’s Law at first interface,

$\displaystyle \frac{sin60}{sin{{r}_{1}}}=\sqrt{3}$

$\displaystyle sin{{r}_{1}}=\frac{sin60}{\sqrt{3}}=\frac{1}{2}$

Now deviation,

$\displaystyle \delta ={{i}_{1}}+{{i}_{2}}-A$

$\displaystyle \Rightarrow {{60}^{\circ }}={{60}^{\circ }}+{{i}_{2}}-A$

$\displaystyle \Rightarrow {{i}_{2}}=A$

Now applying Snell’s Law at second interface,

$\displaystyle \frac{sin{{r}_{2}}}{sin{{i}_{2}}}=\frac{1}{\sqrt{3}}$

$\displaystyle \frac{sin\left( A-{{r}_{1}} \right)}{sinA}=\frac{1}{\sqrt{3}}$

$\displaystyle \frac{sin\left( A-30 \right)}{sinA}=\frac{1}{\sqrt{3}}$

$\displaystyle \frac{\sin A\cos 30-\cos A\sin 30}{\sin A}=\frac{1}{\sqrt{3}}$

$\displaystyle \frac{\sin A\times \frac{\sqrt{3}}{2}-\frac{\cos A}{2}}{\sin A}=\frac{1}{\sqrt{3}}$

$\displaystyle \frac{\sqrt{3}}{2}-\frac{\cot A}{2}=\frac{1}{\sqrt{3}}$

$\displaystyle \cot A=\frac{1}{\sqrt{3}}$

Till now Options (A) and (D) are correct and Option (B) is wrong.

Now for Red Light :

Using prism formula,

$\displaystyle {{\mu }_{red}}=\frac{\sin \left( \frac{A+{{\delta }_{\min }}}{2} \right)}{\sin \frac{A}{2}}=\frac{\sin \left( \frac{60+30}{2} \right)}{\sin \frac{60}{2}}=\frac{\sin 45}{\sin 30}=\frac{1}{\sqrt{2}}\times 2$

Hence option (C) is also correct.

Prism formula | Derivation of prism formula

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