Question

Derive the expression for Angle of prism and angle of deviation in terms of angle of incident, angle of emergence and angle.​

Answer 1

A prism is a transparent medium such as glass bounded by two plane surfaces , inclined to each other at an angle.

We all know,

The deviations produced by the prism depends upon the angle of incidence , refracting angle of prism and the material of prism.

In the attachment , let a monochromatic ray of light PQ be incident on the phase AB.

Thus ,

the equation obtained :

Angle of incidence + angle of emergence  = Angle of prism (A)  + Angle of deviation.

Answer 2

Derivation :-

Refer to the attachment for diagram .

Let's suppose that ;

A = Angle of prism

δ = Angle of deviation

μ1 = refractive index of air (μ1 = 1)

μ2 = μ = refractive index of prism

Now,

At the refracting surface AB :

=> μ1•sin(i1) = μ2•sin(i2)

=> sin(i1) / sin(r1) = μ2 / μ1

=> sin(i1) / sin(r1) = μ / 1

=> sin(i1) / sin(r1) = μ -------(1)

Also,

At the refracting surface AC :

=> μ2•sin(r2) = μ1•sin(r1)

=> sin(i2) / sin(r2) = μ2 / μ1

=> sin(i2) / sin(r2) = μ / 1

=> sin(i2) / sin(r2) = μ --------(2)

Now,

In ∆MAN ,

=> ∠A + ∠AMN + ∠ANM = 180°

=> ∠A + (90° - r1) + (90° - r2) = 180°

=> ∠A - r1 - r2 + 180° = 180°

=> ∠A = r1 + r2 -------(3)

Hence,

The angle of prism is given by ;

∠A = r1 + r2

Now,

In ∆PMN ,

=> δ = ∠PNM + ∠PMN

=> δ = (i1 - r1) + (i2 - r2)

=> δ = i1 + i2 - (r1 + r2)

=> δ = i1 + i2 - ∠A ----------(4)

Hence,

The angle of deviation is given by ;

δ = i1 + i2 - ∠A