Question

in the figure below, a light ray enters a glass slab at point a at incident angle \\( \theta _ { 1 } = 35.8 ^ { \circ } \\) and then undergoes total internal reflection at point b. (the reflection at a is not shown.) what minimum value for the index of refraction of the glass can be inferred from this information?

\\( \bigcirc 1.71 \\)
\\( \bigcirc 1.16 \\)
\\( \bigcirc 1.34 \\)
\\( \bigcirc 1.00 \\)

Explanation:

Step1: Apply Snell's law at point A

Snell's law is \(n_1\sin\theta_1 = n_2\sin\theta_2\). Here, \(n_1 = 1\) (refractive index of air), \(\theta_1=35.8^{\circ}\), and \(n_2 = n\) (refractive index of glass), \(\theta_2\) is the angle of refraction at point A. So, \(\sin\theta_2=\frac{\sin\theta_1}{n}\).

Step2: Analyze the condition for total - internal reflection at point B

For total internal reflection at point B, the critical angle \(\theta_c\) satisfies \(\sin\theta_c=\frac{1}{n}\). Also, from the geometry of the problem, \(\theta_2+\theta_c = 90^{\circ}\), so \(\theta_2 = 90^{\circ}-\theta_c\) and \(\sin\theta_2=\cos\theta_c\).

Step3: Substitute and solve for \(n\)

Since \(\sin\theta_2=\frac{\sin\theta_1}{n}\) and \(\sin\theta_2=\cos\theta_c\), and \(\sin\theta_c=\frac{1}{n}\), using the identity \(\cos\theta_c=\sqrt{1 - \sin^{2}\theta_c}\), we have \(\frac{\sin\theta_1}{n}=\sqrt{1-\frac{1}{n^{2}}}\).
Squaring both sides: \(\frac{\sin^{2}\theta_1}{n^{2}}=1-\frac{1}{n^{2}}\).
Then \(\frac{\sin^{2}\theta_1 + 1}{n^{2}}=1\), and \(n=\sqrt{1+\sin^{2}\theta_1}\).
Substituting \(\theta_1 = 35.8^{\circ}\), \(\sin\theta_1=\sin(35.8^{\circ})\approx0.585\).
\(n=\sqrt{1+(0.585)^{2}}=\sqrt{1 + 0.342}=\sqrt{1.342}\approx1.16\)

Answer:

\(1.16\)