Question

A beam of light of two different wavelengths enters a pane of glass 3.2 mm thick at an angle of incidence of 59°. The indices of refraction for the two different colors are 1.5 and 1.53. Because of dispersion, the colored beams, although parallel, are separated by a small distance. How far apart are they? (Give your answer to the nearest 0.0001 mm)

Answer 1

d = 0.125 mm

Step-by-step explanation:

using snell's law

`(sin\ i)/(sin\ r) = (n_2)/(n_1)`

`r = sin^(-1)((n_1sin\ i )/(n_2))`

n₁ = 1 for air

`r = sin^(-1)((sin\ 59^(\circ) )/(1.5))`

r = 25.09⁰

displacement of color would be

`d = t(sin(i-r))/(cosr)\\d= 3.2(sin(59-25.09))/(cos25.09)\\d= 1.932 mm`

for second color

using snell's law

`(sin\ i)/(sin\ r) = (n_2)/(n_1)`

`r = sin^(-1)((n_1sin\ i )/(n_2))`

n₁ = 1 for air

`r = sin^(-1)((sin\ 59^(\circ) )/(1.53))`

r = 22.58°

displacement of color would be

`d = t(sin(i-r))/(cosr)\\d= 3.2(sin(59-22.58))/(cos22.58)\\d= 2.057 mm`

hence, separation is

d = 2.057 - 1.932

d = 0.125 mm