Worked examples · Sheet 1

Refraction with Snell's law

Theory used: Snell's law Total internal reflection and critical angle Index of refraction Frequency and wavelength in a medium Direction of the deviation Slab with parallel faces

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A ray of yellow light, with wavelength in vacuum $λ = 589 nm$ , passes from air to water ( $n = 1.33$ ) with angle of incidence $θ_{1} = 4 0^{\circ}$ . Compute the angle of refraction and, in water, the frequency, the speed and the wavelength of the light.

sin θ_{2} = \frac{sin θ _{1}}{n}, ν = \frac{c}{λ}, v = \frac{c}{n}, λ_{n} = \frac{λ}{n}

sin θ_{2} ν v λ_{n} = \frac{sin 4 0 ^{\circ}}{1.33} = \frac{0.643}{1.33} \approx 0.483 ⟹ θ_{2} \approx 28. 9^{\circ} = \frac{3.00 \cdot 1 0 ^{8}}{5.89 \cdot 1 0 ^{- 7}} Hz \approx 5.09 \cdot 1 0^{14} Hz = \frac{3.00 \cdot 1 0 ^{8}}{1.33} m/s \approx 2.26 \cdot 1 0^{8} m/s = \frac{589}{1.33} nm \approx 443 nm

The index of air is taken equal to $1.00$ , so Snell’s law $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ reduces to $sin θ_{1} = n sin θ_{2}$ . Water is optically denser than air, hence the ray bends toward the normal, and indeed $28. 9^{\circ} < 4 0^{\circ}$ .

The frequency is imposed by the source and does not change in the passage from one medium to the other, so it is computed with the wavelength in vacuum. The speed in water follows from the definition of the index of refraction, $n = \frac{c}{v}$ . The wavelength in water is the ratio of speed to frequency, that is, $\frac{λ}{n}$ : the light stays yellow, because the color depends on the frequency, and the wavelength shortens in the same ratio as the speed. As a check, $λ_{n} ν = 4.43 \cdot 1 0^{- 7} \cdot 5.09 \cdot 1 0^{14} m/s \approx 2.26 \cdot 1 0^{8} m/s$ , which is the speed found.

Exercise 2

A ray is incident with $θ_{1} = 5 0^{\circ}$ on a glass plate with parallel faces, of thickness $t = 4.0 cm$ , with index $n = 1.50$ , immersed in air. Compute the angle of refraction in the glass, the direction of the emerging ray and its lateral displacement with respect to the incident ray.

The ray crosses the plate from

A

B

and emerges parallel to the incident ray, whose extension is dashed;

d

is the lateral displacement

sin θ_{2} = \frac{sin θ _{1}}{n}, θ_{3} = θ_{1}, d = \frac{t sin ( θ _{1} - θ _{2} )}{cos θ _{2}}

sin θ_{2} d = \frac{sin 5 0 ^{\circ}}{1.50} = \frac{0.766}{1.50} \approx 0.511 ⟹ θ_{2} \approx 30. 7^{\circ} = \frac{4.0 \cdot sin 19. 3 ^{\circ}}{cos 30. 7 ^{\circ}} cm = \frac{4.0 \cdot 0.330}{0.860} cm \approx 1.5 cm

At the first face Snell’s law gives the angle of refraction in the glass, $30. 7^{\circ}$ . The two faces are parallel, so the inner ray strikes the second face at the same angle $θ_{2}$ ; Snell’s law at the second face, $n sin θ_{2} = sin θ_{3}$ , gives $sin θ_{3} = sin θ_{1}$ , hence the ray emerges with $θ_{3} = 5 0^{\circ}$ , parallel to the incident ray.

For the displacement we use the theorem on the slab with parallel faces. The segment $A B$ traveled in the glass has length $\frac{t}{cos θ _{2}} \approx 4.65 cm$ and forms the angle $θ_{1} - θ_{2} = 19. 3^{\circ}$ with the extension of the incident ray; the distance of $B$ from that extension is therefore $A B sin (θ_{1} - θ_{2}) \approx 1.5 cm$ . The result is less than the thickness, as it must be for an angle of incidence less than $9 0^{\circ}$ .

Exercise 3

A glass prism with index $n = 1.50$ has a cross section in the shape of an equilateral triangle, so the apex angle is $A = 6 0^{\circ}$ . A ray lying in the plane of the cross section strikes one face at angle $θ_{1} = 4 5^{\circ}$ , enters the prism and exits through the second face. Compute the exit angle $θ_{4}$ and the angle of deviation $δ$ between the outgoing ray and the incident ray.

sin θ_{2} = \frac{sin θ _{1}}{n}, θ_{3} = A - θ_{2}, sin θ_{4} = n sin θ_{3}, δ = θ_{1} + θ_{4} - A

sin θ_{2} θ_{3} sin θ_{4} δ = \frac{0.707}{1.50} \approx 0.471 ⟹ θ_{2} \approx 28. 1^{\circ} = 6 0^{\circ} - 28. 1^{\circ} = 31. 9^{\circ} = 1.50 \cdot sin 31. 9^{\circ} \approx 0.792 ⟹ θ_{4} \approx 52. 4^{\circ} = 4 5^{\circ} + 52. 4^{\circ} - 6 0^{\circ} \approx 37. 4^{\circ}

At the first face Snell’s law gives the angle $θ_{2}$ of the inner ray with the normal to that face. To find the angle of incidence $θ_{3}$ on the second face the geometry of the prism is needed: the two normals form with each other an angle equal to $A$ , because they are perpendicular to two faces that form the angle $A$ ; in the triangle that the inner ray forms with the two normals, the exterior angle is therefore $A = θ_{2} + θ_{3}$ .

At the second face the light passes from glass to air. The glass-air critical angle is $arcsin \frac{1}{1.50} \approx 41. 8^{\circ}$ and $θ_{3} = 31. 9^{\circ}$ is smaller, hence the ray exits, bending away from the normal, with $θ_{4} \approx 52. 4^{\circ}$ . Each face deviates the ray toward the base of the prism, the first by $θ_{1} - θ_{2}$ and the second by $θ_{4} - θ_{3}$ ; adding and using $θ_{2} + θ_{3} = A$ gives the total deviation $δ = θ_{1} + θ_{4} - A \approx 37. 4^{\circ}$ .