Problem 14 - University Physics with Modern Physics

Question

A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 726 \(\mathrm{nm}\) and its wavelength in the glass is 544 \(\mathrm{nm} .\) If the ray in water makes an angle of \(42.0^{\circ}\) with respect to the normal to the interface, what angle does the refracted ray in the glass make withrespect to the normal?

Step-by-Step Solution

Answer

The angle of refraction in the glass is approximately \(30^{\circ}\).

Step 1: Identify Refractive Indices

To find the angle of refraction, we first need to determine the refractive indices of water and glass. The refractive index is the ratio of the wavelength in vacuum (or air) to the wavelength in the medium. For water: \( n_{water} = \frac{\lambda_{vacuum}}{\lambda_{water}} = \frac{726}{726} = 1.00 \). For glass: \( n_{glass} = \frac{\lambda_{vacuum}}{\lambda_{glass}} = \frac{726}{544} \approx 1.3353 \).

Step 2: Apply Snell's Law

According to Snell's Law, the relationship between the angles of incidence and refraction and the refractive indices is given by \( n_{water} \sin(\theta_{water}) = n_{glass} \sin(\theta_{glass}) \). Plugging in the known values, we have: \( 1.00 \cdot \sin(42^{\circ}) = 1.3353 \cdot \sin(\theta_{glass}) \).

Step 3: Solve for Angle of Refraction

To find \( \theta_{glass} \), rearrange Snell's Law to solve for \( \sin(\theta_{glass}) \): \( \sin(\theta_{glass}) = \frac{\sin(42^{\circ})}{1.3353} \). Calculate \( \sin(42^{\circ}) \approx 0.6691 \), then \( \sin(\theta_{glass}) = \frac{0.6691}{1.3353} \approx 0.5010 \). Now find \( \theta_{glass} \) by taking the inverse sine: \( \theta_{glass} = \arcsin(0.5010) \approx 30^{\circ} \).

Key Concepts

Refractive Index

The refractive index (often symbolized as "n") is a fundamental concept in optics, describing how light propagates through different media. It is a unitless number that indicates how much slower light travels in a medium compared to a vacuum.

• It is calculated as the ratio of the speed of light in a vacuum to its speed in the medium.
• This can also be expressed as the ratio of the wavelength of light in a vacuum to its wavelength in the medium.
• For example, in the given problem, the refractive index of water was considered as 1.00 because the wavelength of light in water matches its original wavelength, assuming it to be vacuum.
• Glass, however, has a refractive index of approximately 1.3353, as calculated from its changed wavelength.

Understanding refractive indices helps us to apply Snell's Law, which governs how much light bends when it moves from one medium to another.

Wavelength

Wavelength is a critical parameter in understanding light behavior, especially in different media. It's the distance between successive crests of a wave, usually measured in nanometers (nm) for light.

• Light's wavelength can vary significantly between different mediums, as the speed of light changes while the frequency remains constant.
• In this exercise, the wavelength is 726 nm in water and 544 nm in glass. This change in wavelength is crucial for calculating the refractive index.
• The shorter wavelength in the glass relates directly to the higher refractive index, indicating slower light speed within the glass.

These variations in wavelength cause the bending effect we measure and calculate using Snell's Law, determining exactly how light behaves at the boundary between two materials.

Angle of Refraction

The angle of refraction is essential for predicting how light bends when transitioning between different mediums. Light changes direction at the interface of two states of matter due to differences in refractive indices.

• Measured with respect to the normal (an imaginary line perpendicular to the surface), this angle helps us determine the new path of the refracted ray.
• Snell's Law mathematically describes the relationship between the angle of incidence and angle of refraction: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \).
• In the problem, we calculated it by first determining the sine of the angle in water and using the refractive indices to find the sine of the angle in glass.
• This process ultimately yielded an angle of refraction in glass of approximately 30°. It helps us visualize how light bends inward when moving from a less dense to a denser medium like glass.

Comprehending how light refracts is essential for applications across various fields, from designing lenses to understanding and creating optical devices.