Question

a beam of light strikes an air/water surface. the water has an index of refraction of 1.33. the angle of incidence is 83.0 degrees. what is the angle of refraction in the water? 45.2 degrees 83.0 degrees 41.4 degrees 48.3 degrees

Explanation:

Step1: Apply Snell's Law

Snell's Law is \(n_1\sin\theta_1 = n_2\sin\theta_2\). Here, \(n_1 = 1\) (refractive index of air), \(\theta_1=83.0^{\circ}\), \(n_2 = 1.33\) (refractive index of water). So, \(\sin\theta_2=\frac{n_1\sin\theta_1}{n_2}\).

Step2: Substitute values

Substitute \(n_1 = 1\), \(\theta_1 = 83.0^{\circ}\), \(n_2=1.33\) into the formula. \(\sin\theta_2=\frac{1\times\sin(83.0^{\circ})}{1.33}\). Since \(\sin(83.0^{\circ})\approx0.9925\), then \(\sin\theta_2=\frac{0.9925}{1.33}\approx0.7463\).

Step3: Find \(\theta_2\)

Take the inverse - sine of \(0.7463\). \(\theta_2=\sin^{- 1}(0.7463)\approx48.3^{\circ}\)

Answer:

48.3 degrees