Question:

A ray of light travels from medium A into medium B, as shown in the diagram below. The light ray is incident on the interface at an angle of 60° with the normal line. The indices of refraction for medium A and medium B are 1.4 and 1.6, respectively. Answer the following questions:

a) Calculate the angle of refraction for the light ray at the interface. b) Determine whether the light ray undergoes total internal reflection at the interface. c) If the light ray undergoes refraction, calculate the angle of deviation upon exiting medium B.

Answer:

a) To calculate the angle of refraction, we can use Snell's law, which states:

where:

• $\theta_1$ is the angle of incidence
• $\theta_2$ is the angle of refraction
• $n_1$ is the refractive index of the initial medium (medium A)
• $n_2$ is the refractive index of the final medium (medium B)

In this case, we have:

• $\theta_1 = 60°$
• $n_1 = 1.4$
• $n_2 = 1.6$

Substituting these values into Snell's law:

Solving for $\theta_2$:

Using the inverse sine function ($\sin^{-1}$) to find $\theta_2$:

Therefore, the angle of refraction for the light ray at the interface is approximately 49.3°.

b) To determine whether the light ray undergoes total internal reflection, we need to compare the angle of incidence with the critical angle. The critical angle is the angle of incidence that would result in an angle of refraction of 90°.

The critical angle can be calculated using the equation:

In this case:

Since the sine of an angle cannot be greater than 1, it means that the angle of incidence (60°) is less than the critical angle. Therefore, the light ray does not undergo total internal reflection at the interface.

c) Since the light ray undergoes refraction, we can calculate the angle of deviation using the following formula:

Substituting the known values:

Therefore, the angle of deviation upon exiting medium B is approximately 10.7°.