Question

a narrow beam of light containing red (λr = 660 nm) and blue (λb = 470 nm) wavelengths travels from air through a d = 1.00 cm thick flat piece of crown glass and back to air again. the beam strikes at a θ = 30.0° incident angle. refractive index of red and blue are nr = 1.509 and nb = 1.517. (a) by what angle are the red and blue separated when they emerge? (b) what is the ratio of velocities of two colors? (c) find the path length difference of two colors?

Explanation:

Step1: Find the refracted angles for red and blue

Using Snell's law \(n_1\sin\theta_i = n_2\sin\theta_r\). Given \(n_1 = 1\) (air), \(\theta_i=30.0^{\circ}\), \(n_{R}=1.509\) and \(n_{B}=1.517\).
For red: \(\sin\theta_{rR}=\frac{n_1\sin\theta_i}{n_{R}}=\frac{1\times\sin30^{\circ}}{1.509}\), \(\theta_{rR}=\sin^{- 1}(\frac{0.5}{1.509})\approx19.2^{\circ}\)
For blue: \(\sin\theta_{rB}=\frac{n_1\sin\theta_i}{n_{B}}=\frac{1\times\sin30^{\circ}}{1.517}\), \(\theta_{rB}=\sin^{- 1}(\frac{0.5}{1.517})\approx19.0^{\circ}\)
When the light emerges back into air, using Snell's law again \(n_2\sin\theta_{r}=n_1\sin\theta_{e}\). Since \(n_1 = 1\) (air - exit medium) and \(n_2\) is the refractive index of glass. The angle of emergence \(\theta_{eR}\) and \(\theta_{eB}\) satisfy \(\theta_{eR}=\theta_i\) (by symmetry of Snell's law for parallel - sided slab) in terms of the relationship between the incident and exit angles. The separation angle \(\Delta\theta=\theta_{eR}-\theta_{eB}\). But we can also calculate the lateral displacement difference. However, another way is to use the formula for the deviation of light through a parallel - sided slab. The deviation \(\delta = 2(\theta_i-\theta_r)\) (for small angles approximation, but we can calculate exactly). The actual separation angle when they emerge:
The angle of refraction at the first interface for red \(\theta_{rR}\) and for blue \(\theta_{rB}\). When the light exits, the angle of emergence \(\theta_{eR}\) and \(\theta_{eB}\) (using Snell's law \(n_{R}\sin\theta_{rR}=n_1\sin\theta_{eR}\) and \(n_{B}\sin\theta_{rB}=n_1\sin\theta_{eB}\)). Since \(n_1 = 1\), \(\theta_{eR}=\sin^{-1}(n_{R}\sin\theta_{rR})\) and \(\theta_{eB}=\sin^{-1}(n_{B}\sin\theta_{rB})\). But by the property of parallel - sided slab, the incident and exit angles (with respect to the normal) are equal. The separation angle comes from the difference in the refraction angles inside the glass.
The formula for the lateral displacement \(y = d\frac{\sin(\theta_i-\theta_r)}{\cos\theta_r}\). The difference in lateral displacement \(\Delta y\) leads to a separation angle. But a more straightforward way (using the fact that for a parallel - sided slab, the emerging ray is parallel to the incident ray in terms of the direction in the external medium). The separation is due to the different refraction angles inside the glass.
The angle between the two emerging rays:
We know that when the light enters the glass, \(n_1\sin\theta_i=n_{R}\sin\theta_{rR}\) and \(n_1\sin\theta_i=n_{B}\sin\theta_{rB}\). When it exits, \(n_{R}\sin\theta_{rR}=n_1\sin\theta_{eR}\) and \(n_{B}\sin\theta_{rB}=n_1\sin\theta_{eB}\). The separation angle \(\Delta\theta\) (approximate, since the rays are parallel to the incident ray in the air) is \(0\) (because of the parallel - sided slab property, the incident and exit angles (with respect to the normal) are equal for each color). But actually, we made a mistake above. Let's use the formula for the deviation of light through a parallel - sided slab. The deviation \(\delta\) of a ray passing through a parallel - sided slab of thickness \(d\) is given by \(\delta=\theta_i-\theta_r\) (for each color). The two rays (red and blue) have different \(\theta_r\).
\(\theta_{rR}=\sin^{-1}(\frac{\sin\theta_i}{n_{R}})=\sin^{-1}(\frac{\sin30^{\circ}}{1.509})\approx19.2^{\circ}\), \(\theta_{rB}=\sin^{-1}(\frac{\sin30^{\circ}}{1.517})\approx19.0^{\circ}\)
The separation angle \(\Delta\theta = 2(\theta_{rB}-\theta_{rR})\) (using the geometry of the slab, the total deviation for each ray is \(2(\theta_i - \theta_r)\) and the separation is the difference in their deviations). \(\Delta\theta=2(\sin^{-1}(\frac{\sin30^{\circ}}{1.509})-\sin^{-1}(\frac{\sin30^{\circ}}{1.517}))\approx2(19.2 - 19.0)=0.4^{\circ}\)

Step2: Calculate the ratio of velocities

The refractive index \(n=\frac{c}{v}\), so \(v=\frac{c}{n}\). The ratio of velocities \(\frac{v_{R}}{v_{B}}=\frac{n_{B}}{n_{R}}\)
Substituting \(n_{R}=1.509\) and \(n_{B}=1.517\), \(\frac{v_{R}}{v_{B}}=\frac{1.517}{1.509}\approx1.0053\)

Step3: Find the path - length difference

The path length inside the glass for a ray with refraction angle \(\theta_r\) and thickness \(d\) is \(L=\frac{d}{\cos\theta_r}\)
For red: \(L_{R}=\frac{d}{\cos\theta_{rR}}\), \(\theta_{rR}=\sin^{-1}(\frac{\sin30^{\circ}}{1.509})\), \(\cos\theta_{rR}=\sqrt{1 - (\frac{\sin30^{\circ}}{1.509})^2}=\sqrt{1-\frac{0.25}{2.277}}\approx0.945\), \(L_{R}=\frac{1.00\space cm}{0.945}\approx1.06\space cm\)
For blue: \(L_{B}=\frac{d}{\cos\theta_{rB}}\), \(\theta_{rB}=\sin^{-1}(\frac{\sin30^{\circ}}{1.517})\), \(\cos\theta_{rB}=\sqrt{1 - (\frac{\sin30^{\circ}}{1.517})^2}=\sqrt{1-\frac{0.25}{2.301}}\approx0.943\), \(L_{B}=\frac{1.00\space cm}{0.943}\approx1.06\space cm\)
The path - length difference \(\Delta L = L_{B}-L_{R}\)
\(L_{R}=\frac{d}{\sqrt{1 - (\frac{\sin\theta_i}{n_{R}})^2}}\), \(L_{B}=\frac{d}{\sqrt{1 - (\frac{\sin\theta_i}{n_{B}})^2}}\)
\(L_{R}=\frac{1}{\sqrt{1 - (\frac{0.5}{1.509})^2}}\space cm\approx1.06\space cm\), \(L_{B}=\frac{1}{\sqrt{1 - (\frac{0.5}{1.517})^2}}\space cm\approx1.06\space cm\)
\(\Delta L=\frac{d}{\sqrt{1 - (\frac{\sin\theta_i}{n_{B}})^2}}-\frac{d}{\sqrt{1 - (\frac{\sin\theta_i}{n_{R}})^2}}\)
\(\Delta L=\frac{1}{\sqrt{1 - (\frac{0.5}{1.517})^2}}-\frac{1}{\sqrt{1 - (\frac{0.5}{1.509})^2}}\approx0.002\space cm = 2\times10^{-5}\space m\)

Answer:

a. The separation angle is approximately \(0.4^{\circ}\)
b. The ratio of velocities \(\frac{v_{R}}{v_{B}}\approx1.0053\)
c. The path - length difference is approximately \(2\times 10^{-5}\space m\)