Question

Light hits the surface of water at an incident angle of \(30.0^{\circ}\) with respect to the normal to the surface. What is the angle between the reflected ray and the refracted ray?

Short answer

Answer: The angle between the reflected ray and the refracted ray is approximately 7.91°.

Step-by-step solution

Find the angle of reflection

According to the law of reflection, the angle of incidence is equal to the angle of reflection. Therefore, the angle of reflection is \(30.0^{\circ}\).

Use Snell's law to find the angle of refraction

Snell's law states that: \(n_1\sin(\theta_1) = n_2\sin(\theta_2)\) where \(n_1\) and \(n_2\) are the refractive indices of the two media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively. In our case, \(\theta_1 = 30^{\circ}\). The refractive index of air is approximately 1, and the refractive index of water is approximately 1.33. We can plug in the values to find the angle of refraction: \(1\sin(30^{\circ}) = 1.33\sin(\theta_2)\) \(\sin(\theta_2) = \dfrac{1\cdot\sin(30^{\circ})}{1.33}\) \(\sin(\theta_2) = \dfrac{0.5}{1.33}\) Now, we can find the angle: \(\theta_2 = \arcsin\left(\dfrac{0.5}{1.33}\right)\) \(\theta_2 \approx 22.09^{\circ}\)

Calculate the angle between the reflected ray and the refracted ray

Now that we have the angle of reflection and the angle of refraction, we can find the angle between the reflected ray and the refracted ray by subtracting the angle of refraction from the angle of reflection: \(\text{Angle between reflected and refracted rays} = \theta_{\text{reflection}} - \theta_{\text{refraction}}\) \(= 30^{\circ} - 22.09^{\circ}\) \(= 7.91^{\circ}\) Hence, the angle between the reflected ray and the refracted ray is \(7.91^{\circ}\).