Question

the critical angle for a ray incident in material x at the boundary of material x and material y is found to be 58.8 degrees. if the index of refraction for material y is 1.06, what is the index of refraction of material x, given that light is going from material y to x and x has a higher refractive index? 1.86 1.24 1.03 this situation is impossible.

Explanation:

Step1: Recall the formula for critical angle

The formula for critical angle \( \theta_c\) when light is going from a medium with refractive index \(n_1\) to \(n_2\) (\(n_1>n_2\)) is \( \sin\theta_c=\frac{n_2}{n_1}\). Here, light is going from \(Y\) to \(X\), so \(n_Y\) is \(n_2\) and \(n_X\) is \(n_1\), and \(\theta_c = 58.8^{\circ}\), \(n_Y = 1.06\).

Step2: Rearrange the formula to solve for \(n_X\)

From \( \sin\theta_c=\frac{n_Y}{n_X}\), we can rearrange it to \(n_X=\frac{n_Y}{\sin\theta_c}\).
Substitute \(n_Y = 1.06\) and \(\theta_c = 58.8^{\circ}\) (so \(\sin\theta_c=\sin(58.8^{\circ})\approx0.855\)) into the formula: \(n_X=\frac{1.06}{0.855}\approx1.24\)

Answer:

1.24