Question: A beam of light is incident on a medium with an angle of incidence of 60°. The refractive index of the medium is 1.5. The beam first refracts and then reflects off a mirror surface situated in the medium. The reflected beam then re-enters the medium. Determine the angles of refraction, reflection, and incidence for the final beam.

Solution: To solve this problem, we will use Snell's law for refraction and the law of reflection.

Let's consider the different stages of the beam's path:

1. Refraction at the first interface (from air to the medium):
   Using Snell's law:
   $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$
   where
   $n_1$ = refractive index of the first medium (air) = 1
   $n_2$ = refractive index of the second medium = 1.5
   $\theta_1$ = angle of incidence = 60° (given)
   $\theta_2$ = angle of refraction (to be determined)

Let's substitute the known values and solve for $\sin(\theta_2)$:
$1 \sin(60°) = 1.5 \sin(\theta_2)$
$\sin(60°) = \frac{1.5}{1} \sin(\theta_2)$
$\frac{\sqrt{3}}{2} = \frac{3}{2} \sin(\theta_2)$
$\sin(\theta_2) = \frac{\sqrt{3}}{3}$
$\theta_2 = \sin^{-1}\left(\frac{\sqrt{3}}{3}\right)$
$\theta_2 ≈ 35.26°$

Therefore, the angle of refraction at the first interface is approximately 35.26°.

2. Reflection at the mirror surface: The angle of incidence and the angle of reflection are equal. Therefore, the angle of reflection at the mirror surface will also be 35.26°.

3. Refraction at the second interface (from the mirror surface back into the medium): Using Snell's law again:
   $n_1 \sin(\theta_3) = n_2 \sin(\theta_4)$
   where
   $n_1$ = refractive index of the first medium = 1.5
   $n_2$ = refractive index of the second medium (air) = 1
   $\theta_3$ = angle of incidence (equal to the angle of reflection) = 35.26°
   $\theta_4$ = angle of refraction (to be determined)

Let's substitute the known values and solve for $\sin(\theta_4)$:
$1.5 \sin(35.26°) = 1 \sin(\theta_4)$
$1.5 \sin(35.26°) = \sin(\theta_4)$
$\sin(\theta_4) \approx 0.8681$
$\theta_4 = \sin^{-1}(0.8681)$
$\theta_4 ≈ 59.36°$

Therefore, the angle of refraction at the second interface is approximately 59.36°.

In summary, the angles of refraction, reflection, and incidence for the final beam are as follows:

• Angle of refraction (at the first interface) ≈ 35.26°
• Angle of reflection (at the mirror surface) ≈ 35.26°
• Angle of refraction (at the second interface) ≈ 59.36°