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.
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.
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.
\(\theta _{1}=(90-55)^\circ=35^\circ\)
\(n=1.46\)
\(n= \frac{sin\theta_{1}}{sin\theta_{2}}\)
\(1.46= \frac{sin35^\circ}{sin\theta _{2}}\)
\(sin\theta _{2}=0.39286\)
\(\theta _{2}=23.133^\circ\)
\(\theta _{2}=23.1^\circ\)