Question

The refractive index of a prism whose angle A = 60° is `sqrt2`. Then the angle of minimum deviation δm will be ______.

विकल्प

• 60°

• 15°

• 30°

• 45°

Answer 1

The refractive index of a prism whose angle A = 60° is `sqrt2`. Then the angle of minimum deviation δm will be 15°.

Explanation:

Refractive index, `mu = (sin ("A" + (delta"m")/2))/(sin "A"//2)`

Given, A = 60°, `mu = sqrt2`, angle of minimum deviation δm = ?

`mu = (sin (60° + (delta"m")/2))/(sin  ((60°)/2))`

or, `sqrt2 = (sin (60° + (delta"m")/2))/(sin 60°//2)`

`= sin (60° + (delta"m")/2)/(sin 30°)`

`= (sin (60° + (delta "m")/2))/(1//2)`

δm = 30°/2 = 15°