Physics please help!

Question

1. A horizontal beam of light enters a 45o-90o-45o prism at the center of its long side, as shown below. The emerging ray travels in a direction that is 34o below the horizontal. What is the index of refraction of this prism? 

Answer 1

Hi Agnes Optics problems like this are much easier once you sketch out the path the light is taking.

Since the beam enters at the center of the long side (the hypotenuse) of a $45^\circ$-$90^\circ$-$45^\circ$ prism, it hits the surface at a normal angle, so it passes straight through without bending at first. The 'fun' part happens at the second surface.

To find the index of refraction ($n$), we use Snell's Law: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$.

1. Identify the angles: Since the ray exits at $34^\circ$ below the horizontal, and the geometry of this prism gives us a specific incident angle at that second boundary, we use the deviation formula for a prism: $\delta = \theta_1 + \theta_2 - A$.
2. The math: Plugging in the geometry for this specific prism ($A = 45^\circ$ at the exit face) and the deviation of $34^\circ$, we can solve for the internal angle.
3. Snell's Law: Once you have that internal angle, you just plug it into $n = \frac{\sin(\theta_2)}{\sin(\theta_1)}$.

This is a great one! If you have more problems or need help prepping for an exam, just let me know. I'd be happy to help. Good luck!