Problem 10 - Physics

Question

A ray of light impinges from air onto a block of ice \((n=1.309)\) at a \(60.0^{\circ}\) angle of incidence. Assuming that this angle remains the same, find the difference \(\theta_{2, \text { ice }}-\theta_{2, \text { water }}\) water in the angles of refraction when the ice turns to water \((n=1.333)\).

Step-by-Step Solution

Answer

The difference in angles is approximately 0.53°.

Step 1: Understand the Problem

We need to find the difference in the angles of refraction when light passes from air into ice and when the same block has turned into water. The refractive indices for ice and water are given, and we have to use these to calculate both angles.

Step 2: Recall Snell's Law

Snell's Law relates the angle of incidence and the angle of refraction to the refractive indices of the two media. It is given by the formula: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \) where \( n_1 \) is the refractive index of air (approximately 1), \( \theta_1 \) is the angle of incidence, and \( n_2 \) and \( \theta_2 \) are the refractive index and angle of refraction for the second medium.

Step 3: Calculate Angle of Refraction for Ice

Using Snell's Law for ice, set \( n_1 = 1 \), \( \theta_1 = 60.0^{\circ} \), and \( n_2 = 1.309 \) (for ice): \[ \sin \theta_{2,\text{ ice}} = \frac{\sin 60.0^{\circ}}{1.309} \]. Calculate \( \theta_{2,\text{ ice}} \) using \( \sin^{-1}(...) \).

Step 4: Calculate Angle of Refraction for Water

Repeat the calculation with water as the medium, setting \( n_2 = 1.333 \) (for water): \[ \sin \theta_{2,\text{ water}} = \frac{\sin 60.0^{\circ}}{1.333} \]. Calculate \( \theta_{2,\text{ water}} \).

Step 5: Find the Difference

Compute the difference between the two refraction angles: \( \theta_{2,\text{ ice}} - \theta_{2,\text{ water}} \). This difference will tell us how much the angle changes when the block changes from ice to water.

Key Concepts

Angle of Incidence

When light hits a surface, the angle at which it strikes is known as the angle of incidence. This angle is measured from the normal, which is an imaginary line perpendicular to the surface of the medium. In the original exercise, the angle of incidence is given as \(60.0^\circ\). This angle remains constant when transitioning between different mediums, such as air to ice or air to water.
Understanding the angle of incidence is crucial because it helps determine how the light will bend or refract as it enters a new medium. The greater the angle of incidence, the more the light bends. This bending occurs because light travels at different speeds in different materials. Snell's Law uses this angle to bridge the transition between media.

Refractive Index

The refractive index is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. Each material has its own refractive index. For example, in our exercise, ice has a refractive index of approximately 1.309, and water has a refractive index of 1.333.

• A refractive index greater than 1 indicates the medium slows down light compared to a vacuum.
• The refractive index can also be interpreted as how much the light path bends when crossing an interface between two materials.

Snell's Law uses the refractive indices of two different media to calculate how much the light will bend, or refract, as it passes from one medium into another. The equation \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \) shows that the ratio of the sine of the angles of incidence and refraction is equivalent to the ratio of the refractive indices of the corresponding media.

Angle of Refraction

The angle of refraction is the angle between the refracted ray and the normal to the interface. When light enters a new medium, it changes direction, and this change is expressed as the angle of refraction. In our problem, you are asked to calculate this angle for both ice and water.
By utilizing Snell's Law, the angle of refraction can be determined. For ice, the formula is \( \sin \theta_{2,\text{ ice}} = \frac{\sin 60.0^\circ}{1.309} \), and for water, it's \( \sin \theta_{2,\text{ water}} = \frac{\sin 60.0^\circ}{1.333} \). You find \( \theta_{2,\text{ ice}} \) and \( \theta_{2,\text{ water}} \) using these formulas.
The angle of refraction is essential for understanding light behavior in different media. It provides insight into optical phenomena like bending and focusing, crucial for applications in lenses and other optical technologies.