A prism in the form of an equilateral triangle (below) has a refractive index \(n=1.6\). A beam of light is incident from air onto the prism at \(\theta_1=50^{\circ}\). a) What is \(\theta_2\) ? b) What is \(\theta_4\) ? c) How can the setup be changed so that total internal reflection takes place? d) A thin film of magnesium fluoride \((n=1.38)\) is painted on the left side of the prism to reduce reflection of red light \((\lambda=630 \mathrm{~nm})\) at normal incidence. What is the wavelength of the red light inside the film? e) What should be its thickness to minimize the reflection of red light? f) If the light is glancing off the prism at an angle, rather than at normal incidence, do you expect the reduction in the reflection of light to be more effective at longer or shorter wavelengths, or to remain the same?

Step by step solution

01

Snell's Law

Use Snell's Law to determine the angle of refraction, \(\theta_2\): \(n_1 \cdot \sin{\theta_1} = n_2 \cdot \sin{\theta_2}\), where \(n_1 = 1\) (refractive index of air), \(n_2=1.6\) (refractive index of prism), and \(\theta_1 = 50^{\circ}\).

02

Solve for \(\theta_2\)

Solve for \(\theta_2\): \(1 \cdot \sin{50^{\circ}} = 1.6 \cdot \sin{\theta_2}\); \(\sin{\theta_2} = \frac{\sin{50^{\circ}}}{1.6}\); \(\theta_2 = \arcsin{\frac{\sin{50^{\circ}}}{1.6}} \approx 18.63^{\circ}\). b) Find \(\theta_4\):

03

Geometry

Use the triangle geometry of the prism to find that the interior angles of an equilateral triangle sum up to \(180^{\circ}\).

04

Angle \theta_3

Since the prism is equilateral, there exists an angle \(\theta_3\) such that \(\theta_3 = 60^{\circ} - \theta_2\) (due to the deviation of the angles inside an equilateral triangle).

05

Snell's Law again

Use Snell's Law again to find \(\theta_4\): \(n_1 \cdot \sin{\theta_3} = n_2 \cdot \sin{\theta_4}\), where \(n_1 = 1.6\), \(n_2=1\), and \(\theta_3 = 60^{\circ} - \theta_2\).

06

Solve for \(\theta_4\)

Solve for \(\theta_4\): \(1.6 \cdot \sin{(60^{\circ}-\theta_2)} = 1 \cdot \sin{\theta_4}\); \(\sin{\theta_4} = 1.6 \cdot \sin{(60^{\circ}-\theta_2)}\); \(\theta_4 = \arcsin{[1.6 \cdot \sin{(60^{\circ}-18.63^{\circ})}]} \approx 65.6^{\circ}\). c) Total internal reflection:

07

Calculate critical angle

Calculate the critical angle from air to the prism (\(n_1<n_2\)): \(\theta_c = \arcsin{\frac{n_1}{n_2}} = \arcsin{\frac{1}{1.6}} \approx 38.21^{\circ}\).

08

Modify the setup

For total internal reflection to take place, the angle of incidence \(\theta_1\) should be greater than the critical angle \(\theta_c\). So, the setup should be modified such that the angle of incidence is greater than \(38.21^{\circ}\). d) Wavelength of red light inside the film:

09

Calculate wavelength

Use the formula \(\lambda_{inside}=\frac{\lambda_{outside}}{n}\). The refractive index of magnesium fluoride is \(n=1.38\), and the wavelength of red light is \(\lambda=630\,\mathrm{nm}\). \(\lambda_{inside}= \frac{630\,\mathrm{nm}}{1.38} \approx 456\,\mathrm{nm}\). e) Optimal thickness to minimize reflection:

10

Constructive interference formula

Use the constructive interference formula to find the optimal thickness of the film: \(2nt = (m + \frac{1}{2})\lambda\), with \(n = 1.38\), \(\lambda = 456\,\mathrm{nm}\), and \(m=0\) for the first minimum.

11

Calculate thickness

Solve for the thickness \(t\): \(2(1.38)t = (\frac{1}{2})(456\,\mathrm{nm})\); \(t = \frac{456\,\mathrm{nm}}{4(1.38)} \approx 82.6\,\mathrm{nm}\). f) Changes in wavelength due to glancing angle:

12

Qualitative analysis

The reduction in the reflection of light would be more effective at shorter wavelengths due to the dispersion of light, as shorter wavelengths tend to have a higher refractive index than longer wavelengths.

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