What is the Minimum Deviation Angle of a Prism with an Apex Angle of 50 Degrees?

[JSGandora]

Homework Statement

The apex angle of a prism is $50.0^\circ$ and the index of refraction of the prism material is $1.66$. What is the minimum deviation angle?

Homework Equations

Snell's Law: $n_1\sin(\theta_1)=n_2\sin(\theta_2)$

The Attempt at a Solution

How would you approach this? I get a deviation angle of $50+\arcsin(1.66\sin(\theta))+\arcsin\left(\frac{ \sin(50-1.66 \sin(\theta))}{1.66}\right)$ using Snell's Law and angle chasing but I don't know how to minimize that. $\theta$ is the initial angle of incidence.

For some reason I think the solution should be much simpler than that. Can someone help me?

 

[anathema]

Hello, thank you for your question.

To find the minimum deviation angle, we can use the formula for minimum deviation, which is given by:

\delta_{min}=A-2\theta_i

Where A is the apex angle and \theta_i is the angle of incidence.

Substituting the given values, we get:

\delta_{min}=50-2\theta_i

To minimize this angle, we need to find the value of \theta_i that will result in the smallest value for \delta_{min}. This can be achieved by taking the derivative of \delta_{min} with respect to \theta_i and setting it equal to 0.

\frac{d\delta_{min}}{d\theta_i}=-2=0

Solving for \theta_i, we get:

\theta_i=\frac{50}{2}=25

Therefore, the minimum deviation angle is:

\delta_{min}=50-2(25)=0

I hope this helps. Let me know if you have any further questions.