Question

now do it again, but this time on your own, with a slightly different triangle apex (top) at and your own unique index of refraction.
a prism (n = 1.36) with an apex angle of 50° has a beam of light strike it at 35° fro normal. find the angle of deviation. use the previous example as the steps will be the sa

Explanation:

Step1: Apply Snell's Law at the first surface

Snell's Law is \(n_1\sin\theta_1 = n_2\sin\theta_2\). Here, \(n_1 = 1\) (air), \(\theta_1=35^{\circ}\), and \(n_2 = 1.36\). So, \(\sin\theta_2=\frac{n_1\sin\theta_1}{n_2}=\frac{1\times\sin35^{\circ}}{1.36}\approx\frac{0.574}{1.36}\approx0.422\), and \(\theta_2\approx25^{\circ}\) (using \(\sin^{- 1}\) function).

Step2: Find the angle of incidence at the second surface

The apex angle \(A = 50^{\circ}\). Using the relation \(A=\theta_2+\theta_3\), we get \(\theta_3=A - \theta_2\). Substituting \(A = 50^{\circ}\) and \(\theta_2 = 25^{\circ}\), we have \(\theta_3=25^{\circ}\)

Step3: Apply Snell's Law at the second surface

Again using \(n_1\sin\theta_1 = n_2\sin\theta_2\), now \(n_1 = 1.36\), \(\theta_1=\theta_3 = 25^{\circ}\), and \(n_2 = 1\). So, \(\sin\theta_2=1.36\times\sin25^{\circ}\approx1.36\times0.423\approx0.575\), and \(\theta_2\approx35^{\circ}\) (using \(\sin^{-1}\) function)

Step4: Calculate the angle of deviation \(\delta\)

The formula for the angle of deviation of a prism is \(\delta=\theta_{i1}+\theta_{i2}-A\). Here, \(\theta_{i1} = 35^{\circ}\), \(\theta_{i2}\approx35^{\circ}\), and \(A = 50^{\circ}\). So, \(\delta=35^{\circ}+35^{\circ}-50^{\circ}=20^{\circ}\)

Answer:

The angle of deviation is \(20^{\circ}\)