Question

Consider a configuration of n identical units, each consisting of three layers. The first layer is a column of air of height h = `1/3` cm, and the second and third layers are of equal thickness d = `(sqrt3 - 1)/2` cm and refractive indices μ₁ = `sqrt(3/2)` and μ₂ = `sqrt3` respectively.

A light source O is placed on the top of the first unit, as shown in the figure. A ray of light from O is incident on the second layer of the first unit at an angle of θ = 60° to the normal. For a specific value of n, the ray of light emerges from the bottom of the configuration at a distance l = `8/sqrt3` cm, as shown in the figure. The value of n is ...........

Answer 1

tan 60° = `x_1/((1/3))`

x₁ = `1/sqrt3` cm

sin 60° = `sqrt(3/2)` sin θ₁

`sqrt3/2 = sqrt(3/2) sin θ_1`

⇒ θ₁ = 45°

Now, tan θ₁ = `x_2/d`

x₂ = d tan θ₁

= `((sqrt3 - 1)/2)` tan 45°

= `((sqrt3 - 1)/2)` cm

Again, by Snell’s law

µ₁ sin θ₁ = µ₂ sin θ₂

`sin θ_2 = µ_2/µ_1 sin θ_1`

`sin θ_2 = sqrt(3/2)/sqrt3 sin 45°`

= `1/sqrt2 xx 1/sqrt2`

= `1/2`

⇒ θ₂ = 30°

tan θ₂ = `x_3/d`

x₃ = d tan θ₂

= `((sqrt3 - 1))/2` tan 30°

= `(1/2 - 1/(2sqrt3))` cm

Total lateral shift by one unit.

x = x₁ + x₂ + x₃

= `1/sqrt3 + ((sqrt3 - 1)/2) + (1/2 - 1/(2sqrt3))`

= `1/sqrt3 - 1/(2sqrt3) + sqrt3/2 - 1/2 + 1/2`

= `(1/sqrt3 - 1/(2sqrt3)) + sqrt3/2`

= `1/(2sqrt3) + sqrt3/2`

= `(1 + (sqrt3)^2)/(2sqrt3)`

= `(1 + 3)/(2sqrt3)`

= `4/(2sqrt3)`

= `2/sqrt3` cm

As there are ‘n’ such units, the net lateral shift = `n xx 2/sqrt3` cm

So, `(2n)/sqrt3 = 8/sqrt3` ...(given)

2n = 8

n = `8/2`

n = 4