Problem 51 - University Physics with Modern Physics

Question

A thin layer of ice \((n=1.309)\) floats on the surface of water \((n=1.333)\) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

Step-by-Step Solution

Answer

(a) \(49.98^\circ\); (b) \(48.75^\circ\).

Step 1: Analyze the Problem

We are asked to find the largest angle with respect to the normal that light can make at the ice-water interface and still pass out into air. This involves using Snell's Law, which relates the angles of incidence and refraction and the refractive indices.

Step 2: Apply Snell's Law for Ice-Air Transition

Snell's Law is given by \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). For light emerging from ice to air, set \( n_1 = 1.309 \) and \( n_2 = 1.0 \) (refractive index of air). The critical angle \( \theta_c \) occurs when \( \theta_2 = 90^\circ \). So, \( \theta_c = \arcsin\left(\frac{n_2}{n_1}\right) = \arcsin\left(\frac{1}{1.309}\right).\)

Step 3: Simplify to Find the Critical Angle for Ice-Air

Solving the equation, \( \theta_c = \arcsin\left(\frac{1}{1.309}\right) \approx 49.98^\circ \). This is the largest angle of incidence at the ice-air interface at which the ray can still exit into the air.

Step 4: Calculate Critical Angle for Water-Air Interface

After the ice melts, we need to find the critical angle at the water-air interface using the same process. Here, \( n_1 = 1.333 \) and \( n_2 = 1.0 \), so \( \theta_c = \arcsin\left(\frac{1}{1.333}\right) \approx 48.75^\circ \).

Key Concepts

Refractive Index

Refractive index is a dimensionless number that describes how fast light travels through a medium compared to the speed of light in a vacuum. It can be viewed as a measure of how much the medium can bend or "refract" light. When light travels from one medium to another, such as from water to air, it changes speed and direction. This change in direction is governed by the refractive indices of the two media involved.

• Each medium has its own refractive index, often denoted by the letter "n".
• For instance, the refractive index of ice is approximately 1.309, while for water, it is 1.333.
• The refractive index of air is usually considered to be 1.0 as light travels the fastest in a vacuum.

Understanding refractive index is crucial for applying Snell's Law, which predicts the path taken by light when it changes medium. The refractive index influences the angles at which light enters or exits a medium.

Critical Angle

The critical angle is the angle of incidence at which light is refracted along the boundary between two media. When the angle of incidence exceeds the critical angle, total internal reflection occurs. This means that instead of passing through the boundary into the second medium, the light is completely reflected back into the first medium.To find the critical angle, we use Snell's Law, where the angle of refraction is set to 90 degrees. This represents the scenario where the light is refracting exactly along the boundary.For the ice-air interface with a refractive index of ice being 1.309, the critical angle can be calculated as follows:\[\theta_c = \arcsin\left(\frac{n_{2}}{n_{1}}\right) = \arcsin\left(\frac{1.0}{1.309}\right) \approx 49.98^\circ\]For the water-air interface, once the ice melts and if the refractive index of water is 1.333, the critical angle is:\[\theta_c = \arcsin\left(\frac{1.0}{1.333}\right) \approx 48.75^\circ\]Both critical angles determine the largest angle at which light can exit the media and enter the air, beyond which total internal reflection will happen.

Light Refraction

Light refraction is the phenomenon where light changes its direction as it passes through different media. This change happens due to a difference in light speed within the two media. The principle of refraction explains many common visual experiences, such as the bending of a straw in a glass of water.When light hits a medium at an angle, it bends towards the normal if it's entering a medium with a higher refractive index, and away from the normal if it's entering a medium with a lower refractive index. Snell's Law mathematically describes this behavior by relating the angles of incidence and refraction to the refractive indices:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]Here, \(n_1\) and \(n_2\) are the refractive indices of mediums 1 and 2 respectively, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction.

• When light travels from ice (n = 1.309) to air (n = 1), the direction of light changes considerably.
• The degree of light bending is determined by the difference in refractive indices.

The concept of light refraction is fundamental in understanding optical devices and natural observations related to light and vision.