Question

A light ray traveling in air (n=1) encounters a medium with an index of refraction of 1.5. The incident angle of the ray is 30 degrees. Determine:

a) The angle of refraction at this interface. b) The critical angle for this interface. c) If the angle of incidence is larger than the critical angle, will the light ray undergo total internal reflection or refraction? If so, at what angle will the ray be reflected? d) If the angle of incidence is smaller than the critical angle, will the light ray undergo total internal reflection or refraction? If so, at what angle will the ray be transmitted?

Answer

a) The angle of refraction at this interface can be found using Snell's Law, which states:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

where $n_1$ and $\theta_1$ are the refractive index and angle of incidence in the first medium, and $n_2$ and $\theta_2$ are the refractive index and angle of refraction in the second medium.

Given that $n_1 = 1 (air)$, $n_2 = 1.5$, and $\theta_1 = 30^\circ$, we can rearrange the equation and solve for $\theta_2$:

$\sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1)$

$\sin(\theta_2) = \frac{1}{1.5} \sin(30^\circ)$

$\sin(\theta_2) \approx 0.3473$

Taking the inverse sine, we find:

$\theta_2 \approx 20.61^\circ$

Therefore, the angle of refraction at this interface is approximately $20.61^\circ$.

b) The critical angle can be calculated using the equation:

$\sin(\theta_c) = \frac{n_2}{n_1}$

$\sin(\theta_c) = \frac{1.5}{1}$

$\sin(\theta_c) = 1.5$

However, there is no angle whose sine is greater than 1, so total internal reflection occurs at this interface.

c) Since the angle of incidence is larger than the critical angle, the light ray will undergo total internal reflection. The angle of reflection can be determined to be the same as the angle of incidence, which is $30^\circ$.

d) Since the angle of incidence is smaller than the critical angle, the light ray will undergo refraction. The angle of transmission can be found using Snell's Law:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_t)$

$1 \sin(30^\circ) = 1.5 \sin(\theta_t)$

$\sin(\theta_t) = \frac{\sin(30^\circ)}{1.5}$

$\sin(\theta_t) \approx 0.2873$

Taking the inverse sine, we find:

$\theta_t \approx 16.72^\circ$

Therefore, the light ray will be transmitted at an angle of approximately $16.72^\circ$.