Problem 47 - University Physics with Modern Physics

Question

Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was somewhat transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of the spot. If the glass plate is 3.10 \(\mathrm{mm}\) thick and the halos have an inner radius of \(5.34 \mathrm{mm},\) what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reflected at the back surface of the plate and returns to the front surface.)

Step-by-Step Solution

Answer

The index of refraction is approximately 1.60.

Step 1: Understand the Problem

We need to determine the index of refraction of the glass plate using the provided measurements. The problem involves total internal reflection within the glass.

Step 2: Identify the Conceptual Framework

The inner radius of the halo is due to the light being totally internally reflected within the glass plate. This involves Snell's Law and the concept of critical angle for total internal reflection.

Step 3: Apply Snell's Law to Determine Critical Angle

Snell's Law is given by \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). For total internal reflection, the angle in the more dense medium (glass) is the critical angle \( \theta_c \). When \( \theta_2 = 90^\circ \), we have \( \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) \), where \( n_2 = 1 \) (air).

Step 4: Calculate the Tangent of the Critical Angle

The angle corresponding to the inner halo radius can be seen as forming a tangent with the critical angle when considering the height (thickness of the plate) over the radius. Hence, \( \tan \theta_c = \frac{3.10 \, \text{mm}}{5.34 \, \text{mm}} \).

Step 5: Solve for the Critical Angle

Using \( \tan \theta_c = \frac{3.10}{5.34} \), find \( \theta_c \) by calculating \( \theta_c = \tan^{-1}\left(\frac{3.10}{5.34}\right) \).

Step 6: Determine the Index of Refraction

Using the critical angle from the previous step, solve for the index of refraction using \( n_1 = \frac{1}{\sin \theta_c} \). Calculate this value to find the index of refraction.

Key Concepts

Total Internal Reflection

Total internal reflection is a fascinating phenomenon where light waves are completely reflected back into a medium, rather than passing through to another. This occurs under specific conditions that involve two main factors: the angle of incidence and the indices of refraction of the two media.

When light travels from a medium with a higher index of refraction (like glass) to one with a lower index of refraction (like air), it bends away from the normal. If the angle at which the light hits the boundary between the two media is larger than a specific value, known as the critical angle, all of the light will reflect back into the original medium instead of passing through.

This behavior is called total internal reflection. It only occurs when the light is in the denser medium moving towards a less dense medium. One everyday example of total internal reflection is fiber optics, used in many communication technologies.

Snell's Law

Snell's Law is the principle that describes how light bends, or refracts, as it travels between two different media. It is mathematically expressed 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 two media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively. Snell's Law is crucial in calculating the path of light as it moves from one medium to another.

In the context of total internal reflection, we are particularly interested in determining the critical angle. The critical angle occurs in scenarios where \( \theta_2 = 90^\circ \), meaning the refracted light travels along the boundary between the two media. Using the law, the critical angle can be found using:

• \( \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) \)

Understanding Snell's Law helps us predict and manipulate how light behaves, which is essential for designing lenses, prisms, and many optical devices.

Index of Refraction

The index of refraction, often simply called 'refractive index,' is a measure of how much light bends, or refracts, when entering a material. It is a dimensionless number that indicates how fast light travels through a substance compared to the speed of light in a vacuum.

• A higher refractive index means light travels slower in that medium.
• Common materials, like glass and water, typically have refractive indices greater than 1.

The index of refraction is pivotal in the analysis of optical systems. It plays a key role in Snell's Law, where it influences the behavior of light as it crosses boundaries between media. In solving problems like the one involving a photographic plate, determining the refractive index involves understanding the concepts of total internal reflection and employing Snell's Law.

By knowing the refractive index, scientists and engineers can design optical components to control light paths, optimize lenses for cameras, and innovate in fiber optics technology.