Problem statement: The minimum angle of deviation of a prism of refractive index m is equal to its refracting angle. The refractive angle of prism is?

Solution:

Understanding the concept of minimum angle of deviation:
The minimum angle of deviation is the angle between the incident and emergent rays produced by a prism, which passes through the prism symmetrically. The angle of deviation is measured by taking the difference between the angle of incidence and the angle of emergence. The minimum angle of deviation is the smallest angle of deviation that a prism can produce for a particular wavelength.

Deriving the formula for the refracting angle of the prism:
Let A be the refracting angle of the prism. Then the angle of incidence and angle of emergence are both equal to A/2.

Let μ be the refractive index of the prism.

Then, by Snell's law, we have

sin (A/2) / sin [(A+δ)/2] = μ
sin (A/2) / sin [(A-δ)/2] = 1/μ

where δ is the angle of deviation.

Now, using the condition given in the problem that the minimum angle of deviation is equal to the refracting angle, we have

A/2 = δmin

Substituting this in the above equations, we get

sin δmin = 2 sin (A/2) / (μ+1/μ)
cos δmin = (μ-1/μ) / (μ+1/μ)

Using the relation

sin2 θ + cos2 θ = 1

we get

sin2 δmin + cos2 δmin = 1

Substituting the values of sin δmin and cos δmin, we get

4 μ2 sin2 (A/2) + (μ2 - 1)2 = 4 μ2

Simplifying this equation, we get

μ2 = (sin2 (A/2) + 1) / 2 sin2 (A/2)

Now, substituting the value of μ in terms of A, we get

sin2 (A/2) = (1 + sqrt(1 - 1/m2)) / 2m

Taking the square root of both sides, we get

sin (A/2) = sqrt[(1 + sqrt(1 - 1/m2)) / 2m]

Multiplying both sides by 2, we get

sin A = 2 sqrt[(1 + sqrt(1 - 1/m2)) / 2m]

Taking the inverse sine, we get

A = sin-1 {2 sqrt[(1 + sqrt(1 - 1/m2)) / 2m]}

Final Answer: The refracting angle of the prism is A = sin-1 {2 sqrt[(1 + sqrt(1 - 1/m2)) / 2m]}.