Question:

A ray of light passes from air into a medium with a refractive index of 1.5. The incident angle is 45 degrees with respect to the normal. Determine the angle of refraction and the speed of light in the medium.

Answer:

To solve this problem, we can use Snell's law, which relates the angle of incidence to the angle of refraction:

Where:

• $n_1$ and $n_2$ are the refractive indices of the incident and refracted media, respectively.
• $\theta_1$ is the angle of incidence with respect to the normal.
• $\theta_2$ is the angle of refraction with respect to the normal.

Given: $n_1 = 1$ (refractive index of air) $n_2 = 1.5$ (refractive index of the medium) $\theta_1 = 45$ degrees

Let's solve for $\theta_2$ first:

Using a calculator, we find that $\sin(\theta_2) \approx 0.471$. To find $\theta_2$, we can take the inverse sine of this value:

Using a calculator, we find that $\theta_2 \approx 28.1^\circ$.

Now, let's find the speed of light in the medium. The speed of light is given by:

Where:

• $v$ is the speed of light in a medium.
• $c$ is the speed of light in vacuum (approximately $3 \times 10^8$ m/s).
• $n$ is the refractive index of the medium.

Given: $c = 3 \times 10^8$ m/s $n = 1.5$

Calculating this equation, we find that $v \approx 2 \times 10^8$ m/s.

Therefore, the angle of refraction is approximately $28.1^\circ$ and the speed of light in the medium is approximately $2 \times 10^8$ m/s.