Question

A submarine is inspecting the surface of the water with a laser that points from the submarine to the surface of the water and through the air. At what angle will the laser not penetrate the surface of the water but rather reflect entirely back into the water? (Assume \(n_{\text {water }}=1.33\) and \(n_{\text {air }}=1 .\) (A) \(19^{\circ}\) (B) \(29^{\circ}\) (C) \(39^{\circ}\) (D) \(49^{\circ}\)

Short answer

Option (D), \(49^{\circ}\).

Step-by-step solution

- Understand the problem

The problem is asking for the critical angle at which the laser will reflect entirely back into the water when it hits the water-air interface.

- Recall Snell's Law

Snell's Law is given by: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] where \(n_1\) and \(n_2\) are the indices of refraction of the two media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively.

- Determine the critical angle

At the critical angle, \(\theta_c\), the angle of refraction, \(\theta_2\), is \(90^{\circ}\). Thus, the sine of the angle of refraction is 1. We set \(\theta_2\) to \(90^{\circ}\) and rearrange Snell's Law to solve for \(\theta_c\): \[ n_1 \sin(\theta_c) = n_2 \sin(90^{\circ}) \] \[ n_1 \sin(\theta_c) = n_2(1) \] \[ \sin(\theta_c) = \frac{n_2}{n_1} \]

- Substitute the values

Substitute \(n_{\text{water}} = 1.33\) and \(n_{\text{air}} = 1\) into the equation: \[ \sin(\theta_c) = \frac{1}{1.33} \]

- Calculate the critical angle

Calculate \(\theta_c\) using the inverse sine function: \[ \theta_c = \sin^{-1}\left( \frac{1}{1.33} \right) \] Using a calculator, \[ \theta_c \approx 48.8^{\circ} \]

- Choose the closest answer

The closest answer choice to \(48.8^{\circ}\) is \(49^{\circ}\) which corresponds to option (D).

Key concepts

Snell's Law

Snell's Law is a fundamental principle in optics that describes how light bends when it passes between different media. It mathematically expresses the relationship between the angles of incidence and refraction, and the indices of refraction of the two media involved. The law is given by the equation: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] Here, \(n_1\) and \(n_2\) are the indices of refraction of the media, while \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively.

When light travels from a medium with a higher index of refraction to a medium with a lower index of refraction, it bends away from the normal. Conversely, when light travels from a medium with a lower index of refraction to a higher one, it bends toward the normal.

The equation helps us predict the path of light when it transitions between materials such as air, water, or glass. In the context of our exercise, Snell's Law is rearranged to find the critical angle, the angle of incidence where light no longer refracts but reflects completely back into the medium.

Index of Refraction

The index of refraction, denoted by \(n\), is a measure of how much a medium slows down light compared to the speed of light in a vacuum. It can be expressed as: \[ n = \frac{c}{v} \] where \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the medium.

Different materials have different indices of refraction. For instance, the index of refraction for air is approximately 1, meaning light travels almost at its maximum speed. In contrast, water has an index of refraction of about 1.33, indicating that light travels slower through water than through air.

Understanding the index of refraction is crucial for solving problems involving Snell's Law. By knowing the indices of refraction of the two media involved, we can predict the angle at which light will refract or reflect when transitioning between these media. In our exercise, the indices provided are for water (1.33) and air (1.0). These values are used to determine the critical angle.

Total Internal Reflection

Total internal reflection occurs when light attempts to move from a medium with a higher index of refraction to one with a lower index of refraction at an incidence angle greater than the critical angle. Instead of refracting out of the medium, light reflects entirely back into it.

The critical angle, \(\theta_c\), is the angle of incidence at which this phenomenon occurs. It can be calculated using Snell's Law, setting the angle of refraction \(\theta_2\) to \(90^\circ\): \[ n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \] Given that \( \sin(90^\circ) = 1 \), we have: \[ \sin(\theta_c) = \frac{n_2}{n_1} \]

In our exercise, we substitute the given values for water (1.33) and air (1.0) to find: \[ \theta_c = \sin^{-1}\left(\frac{1}{1.33}\right) \approx 48.8^\circ \]

This angle of about 48.8° is the critical angle for light moving from water to air. Hence, any angle of incidence greater than this will result in total internal reflection, causing the laser light to reflect back into the water rather than penetrating the surface.