Question

A light beam strikes a piece of glass at a 62.00 ? 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? Express your answer using two significant figures.

Answer 1

For 450 nm wavelength

Using snell's law:

n1*sin A1 = n2*sin A2

n1 = refractive index of Air = 1

n2 = refractive index of glass = 1.4831

A1 = Angle of incidence = 62 deg

A2 = angle of refraction = ?

A2 = arcsin ((n1/n2)*sin A1)

Using given values:

A2 = arcsin ((1/1.4831)*sin 62 deg) = 36.5368 deg

For 700 nm wavelength

Using snell's law:

n1*sin A1 = n3*sin A3

n1 = refractive index of Air = 1

n3 = refractive index of glass = 1.4754

A1 = Angle of incidence = 62 deg

A3 = angle of refraction = ?

A3 = arcsin ((n1/n3)*sin A1)

Using given values:

A3 = arcsin ((1/1.4754)*sin 62 deg) = 36.7587 deg

So now angle b/w refracted beams = A3 - A2

= 36.7587 - 36.5368 = 0.2219

Angle b/w refracted beams = 0.22 deg

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