Problem 16 - University Physics with Modern Physics

Question

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of \(\theta_{a}=36.2^{\circ},\) the ray refracted into the water makes an angle of \(49.8^{\circ}\) with the normal to the interface. What is the smallest value of the incident angle \(\theta_{a}\) for which none of the ray refracts into the water?

Step-by-Step Solution

Answer

The smallest angle is approximately 62.46° for total internal reflection.

Step 1: Understand the Problem

We have a light ray traveling through glass and hitting the interface between glass and water. We are asked to find the smallest angle of incidence, \(\theta_a\), such that the ray does not refract into the water at all. This occurs when the ray is perfectly reflected, known as total internal reflection.

Step 2: Apply Snell's Law for Given Refracted Angle

Given that the refracted angle \( \theta_w = 49.8^\circ \) and the angle of incidence \( \theta_a = 36.2^\circ \), use Snell's Law: \[ n_g \sin \theta_a = n_w \sin \theta_w \]where \( n_g \) is the index of refraction of the glass and \( n_w \) is the index of refraction of the water.

Step 3: Use Known Indices of Refraction

Typically, the index of refraction for glass is about \( n_g = 1.5 \) and for water \( n_w = 1.33 \). Substitute these values in to find:\[ 1.5 \sin(36.2^\circ) = 1.33 \sin(49.8^\circ) \].Calculate to verify that both sides of the equation are equal.

Step 4: Determine Critical Angle

The critical angle \( \theta_c \) is the angle of incidence at which the refracted angle in the second medium (water) is \( 90^\circ \). Using Snell's Law, set the refracted angle to 90 degrees:\[ n_g \sin \theta_c = n_w \sin(90^\circ) \].Since \( \sin(90^\circ) = 1 \), we find that the critical angle is:\[ \theta_c = \sin^{-1} \left(\frac{n_w}{n_g}\right) = \sin^{-1} \left(\frac{1.33}{1.5}\right) \].

Step 5: Calculate the Critical Angle

Calculate the critical angle:\[ \theta_c = \sin^{-1} \left(\frac{1.33}{1.5}\right) \approx 62.46^\circ \].This is the smallest angle of incidence for total internal reflection, meaning if the incident angle is any smaller, some light would refract into the water.

Key Concepts

Snell's Law

Snell's Law is an essential principle in understanding how light behaves when passing through different media. It explains how light rays bend, or refract, at the boundary between two materials with different indices of refraction.
Follow this simple rule:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
Here, \( n_1 \) and \( n_2 \) are the indices of refraction for the two different media, while \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.

• If the light moves from a medium with a lower index of refraction to a medium with a higher index, the light slows down and bends towards the normal.
• Conversely, if it moves from a higher to a lower index, the light speeds up and bends away from the normal.

In our problem, light transitions from glass (\( n_g = 1.5 \)) to water (\( n_w = 1.33 \)), causing it to refract at an angle of 49.8°, given an incident angle of 36.2° in glass. This application of Snell's Law enables us to predict light's path precisely.

critical angle

The critical angle is a fascinating concept that defines the threshold at which light traveling from a denser to a rarer medium undergoes total internal reflection. This means instead of refracting into the second medium, the light bounces back entirely into the first.
The critical angle can be calculated using the formula derived from Snell's Law:
\[ \theta_c = \sin^{-1} \left( \frac{n_2}{n_1} \right) \]
In this expression, \( n_1 \) is the index of refraction of the initial medium (glass) and \( n_2 \) is the index for the second medium (water). Since the sine of 90° is 1, we substitute this for the refracted angle.
The critical angle is reached when the refracted angle is 90°, meaning the ray runs along the boundary. Calculating for the scenario where light travels from glass to water,
\[ \theta_c = \sin^{-1} \left( \frac{1.33}{1.5} \right) \approx 62.46^\circ \]
This value specifies that an incident angle of less than 62.46° results in some refracted light, while at angles equal to or greater, light is perfectly reflected back.

index of refraction

The index of refraction, also known as the refractive index, reveals how much light slows down when entering a material. Light travels slowest in materials with higher indices. Each medium has its specific index based on its optical density.
The refractive index \( n \) is defined 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.

• A higher index indicates that light travels slower in the material, making glass (\( n = 1.5 \)) denser than water (\( n = 1.33 \)).
• This characteristic determines how much the light bends when entering or leaving a material.

In our scenario, the difference in refractive indices for glass and water causes the light to bend away from the normal as it moves from the glass into the water. The greater the difference in indices, the more pronounced the refraction.