Wavelength and refractive index

We can also use wavelengths to calculate refractive index.

By substituting v_{1}=f\lambda _{1} and v_{2}=f\lambda _{2} into n=\frac{v_{1}}{v_{2}} then:

n=\frac{v_{1}}{v_{2}}=\frac{f\lambda _{1}}{f\lambda _{2}}=\frac{\lambda _{1}}{\lambda _{2}}

Note that the frequency is cancelled out and not needed.

Remember in the medium where the light is faster (ie bigger speed), the angle is bigger and the wavelength is bigger.

n=\frac{v_{1}}{v_{2}}=\frac{\lambda _{1}}{\lambda _{2}}=\frac{sin\theta _{1}}{sin\theta _{2}}

Question

A ray of blue light of frequency 4.8 \times {10^{14}} Hz is incident on the surface of a piece of clear plastic as shown.

A ray of light passes through air at an angle of 55 degrees. Once it hits the plastic, the angle changes slightly. There is a dotted line on normal incidence.

For this light the refractive index of the plastic is 1.48.

Calculate the angle of refraction.

Be careful with the angles given in a question. Here the angle given, 55^\circ, is the angle between the ray and the surface.

Ray passes air at 55 degrees. Hits plastic & angle changes slightly. Angle between 55 degrees and normal incidence is theta 1. Angle between normal incidence and ray of light in plastic is theta 2.

To answer this question you need to use the angle of incidence, that is the angle between the ray and the normal to the surface, \theta _{1}. The angle you need to find is the angle of refraction, \theta _{2}.

\theta _{1}=(90-55)^\circ=35^\circ

n= 1.48

n= \frac{sin\theta_{1}}{sin\theta_{2}}

1.48= \frac{sin35^\circ}{sin\theta _{2}}

sin\theta _{2}=0.3875516

\theta _{2}=22.8022

\theta _{2}= 22.8^\circ

Question

Now calculate the wavelength of the light in air.

v = c = 3.00 \times {10^8}m{s^{ - 1}}

f = 4.8 \times {10^{14}}Hz

v=f\lambda

3.00\times 10^{8}=4.8\times 10^{14}\times \lambda

\lambda =6.25\times 10^{-7}m

Question

Now calculate the wavelength in plastic.

n=1.48

\lambda _{1} = 6.25 \times 10^{-7}m

n=\frac{\lambda _{1}}{\lambda _{2}}

1.48=\frac{6.25\times 10^{-7}}{\lambda _{2}}

\lambda _{2}=4.22\times 10^{-7}m

Wavelength of light in the plastic =4.2 \times 10^{-7}m

Question

The refractive index for red light is 1.46. Calculate the angle of refraction for red light.

\theta _{1}=(90-55)^\circ=35^\circ

n=1.46

n= \frac{sin\theta_{1}}{sin\theta_{2}}

1.46= \frac{sin35^\circ}{sin\theta _{2}}

sin\theta _{2}=0.39286

\theta _{2}=23.133^\circ

\theta _{2}=23.1^\circ