Question

A light beam strikes a piece of glass at a 65.00 ∘ incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4831 and 1.4754, respectively.
What is the angle between the two refracted beams?

Answer 1

Answer:

Explanation:

Lower the refractive index ,  higher the wave length

Refractive index becoming high reduces the velocity and hence wavelength as frequency remains unchanged.

So refractive index  1.4831 will correspond to wavelength of 450 nm.

For  refractive index is 1.4831 , angle of incidence i , angle of refraction r .

Sin i / Sinr = 1.4831

sin65 / sinr = 1.4831

sir r = sin65 / 1.4831

= .9063 / 1.4831

= .6111

r = 37.67 degree

For  refractive index is 1.4754 , angle of incidence i

Sin i / Sinr = 1.4754

sin65 / sinr = 1.4754

sir r = sin65 / 1.4754

= .9063 / 1.4754

= .61427

r = 37.9 degree

angle between two refracted ray

37.9 -37.67

= 0. 23 degree

Answer 2

Answer:

0.2°

Explanation:

Angle of incidence on glass, i = 65°

first wavelength, λ1 = 450 nm

second wavelength, λ1 = 700 nm

first refractive index, μ1 = 1.4831

second refractive index, μ1 = 1.4754

Let the angle of refraction for the first wavelength is r1 and for the second wavelength is r2.

Use Snell's law

$\mu_{1}=\frac{Sin i}{Sin r_{1}}$

$Sin r_{1}=\frac{Sin 65}{1.4831}$

Sin r1 = 0.61109

r1 = 37.7°

$\mu_{2}=\frac{Sin i}{Sin r_{2}}$

$Sin r_{2}=\frac{Sin 65}{1.4754}$

Sin r2 = 0.6143

r2 = 37.9°

Angle between the two refracted beams = r2 - r1 = 37.9° - 37.7° = 0.2°