Question

A thin prism of angle 6.0°, ω = 0.07 and μ_y = 1.50 is combined with another thin prism having ω = 0.08 and μ_y = 1.60. The combination produces no deviation in the mean ray. (a) Find the angle of the second prism. (b) Find the net angular dispersion produced by the combination when a beam of white light passes through it. (c) If the prisms are similarly directed, what will be the deviation in the mean ray? (d) Find the angular dispersion in the situation described in (c).

Answer 1

Given:
For the first prism,
Angle of prism, A' = 6°
Angle of deviation, ω' = 0.07
Refractive index for yellow colour, μ'_y = 1.50
For the second prism,
Angle of deviation, ω = 0.08
Refractive index for yellow colour, μ_y= 1.60
Let the angle of prism for the second prism be A.
The prism must be oppositely directed, as the combination produces no deviation in the mean ray.
(a) The deviation of the mean ray is zero.
Thus, we have:
δ_y = (μ_y – 1)A – (μ'_y – 1)A' = 0
$\therefore$ (1.60 – 1)A = (1.50 – 1)A'
⇒ A = $\frac{0.50\times 6°}{0.60}=5°$



(b) Net angular dispersion on passing a beam of white light:
(μ_y – 1)ωA – (μ_y – 1)ω'A'
⇒ (1.60 – 1)(0.08)(5°) – (1.50 – 1)(0.07)(6°)
⇒ 0.24° – 0.21° = 0.03°



(c) For the prisms directed similarly, the net deviation in the mean ray is given by
δ_y = (μ_y – 1)A + (μ_y – 1)A'
= (1.60 – 1)5° + (1.50 – 1)6°
= 3° + 3° = 6°

(d) For the prisms directed similarly, angular dispersion is given by
δ_v – δ_r = (μ_y – 1)ωA – (μ_y – 1)ω'A'
= 0.24° + 0.21°
= 0.45°