Question

a source of light is in a medium with an index of refraction of 2.29. if the medium on the other side of the surface has an index of 2.15, what is the critical angle?
69.9 degrees
total internal reflection cannot occur.
60.9 degrees
74.9 degrees

Explanation:

Step1: Recall the condition for total internal reflection

Total internal reflection occurs when light travels from a medium with a higher refractive index \(n_1\) to a medium with a lower refractive index \(n_2\). Here, \(n_1 = 2.29\) and \(n_2=2.15\). Since \(n_1>n_2\), total internal reflection is possible.

Step2: Use the formula for critical angle

The formula for the critical angle \(\theta_c\) is \(\sin\theta_c=\frac{n_2}{n_1}\).
Substitute \(n_1 = 2.29\) and \(n_2 = 2.15\) into the formula: \(\sin\theta_c=\frac{2.15}{2.29}\).
Calculate \(\frac{2.15}{2.29}\approx0.939\).
Then \(\theta_c=\sin^{- 1}(0.939)\).
Using a calculator, \(\theta_c\approx69.9^{\circ}\).

Answer:

69.9 degrees