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

Answer:

d = 0.125 mm

Explanation:

using snell's law

$\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}$

$r = sin^{-1}(\dfrac{n_1sin\ i }{n_2})$

n₁ = 1 for air

$r = sin^{-1}(\dfrac{sin\ 59^{\circ} }{1.5})$

r = 25.09⁰

displacement of color would be

$d = t\dfrac{sin(i-r)}{cosr}\\d= 3.2\dfrac{sin(59-25.09)}{cos25.09}\\d= 1.932 mm$

for second color

using snell's law

$\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}$

$r = sin^{-1}(\dfrac{n_1sin\ i }{n_2})$

n₁ = 1 for air

$r = sin^{-1}(\dfrac{sin\ 59^{\circ} }{1.53})$

r = 22.58°

displacement of color would be

$d = t\dfrac{sin(i-r)}{cosr}\\d= 3.2\dfrac{sin(59-22.58)}{cos22.58}\\d= 2.057 mm$

hence, separation is

d = 2.057 - 1.932

d = 0.125 mm