Question

A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is 3/4 th of the angle of prism. Calculate the speed of light in the prism. 

A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.

Answer 1

i = 3/4 A for minimum deviation we know,
A + δ_m = 2i

Therefore, A + δ_m = 2 × 3/4 A

δ_m = (1.5 − 1)A

⇒ δ_m = 0.5 A

For equilateral prism

A = 60° ...(i)

⇒ δ_m = 0.5 × 60° = 30° ...(ii)

Also, for minimum deviation, refractive index of glass w.r.t. air is given by

`∴μ="speed of light in air (c)"/"speed of light in prism"`...(iii)

`thereforemu=sin((A+δ_m)/2)/sin(A/2)`

`∴μ=sin((60+30)^@/2)/sin(60^@/2)`

`∴μ=sin((90^@)/2)/sin(60^@/2)`   ....(iv)

 using  (iii) and (iv)

`∴"speed of light in prism"="speed of light in air (c)"/μ`

`∴"speed of light in prism" =sin(60^@/2)/sin(90^@/2) xx c`

`= (1/2)/(1/sqrt2)xx  3 × 10^8`

`= 3/sqrt2 × 10^8 "m/s"`

`= 2.12 × 10^8 "m/s"`