Question

A ray of light \(\left(f=5 \times 10^{14} \mathrm{Hz}\right)\) travels from air into crystal into chromium. If the indices of refraction of air, crystal, and chromium are \(1,2,\) and \(3,\) respectively, and the incident angle is \(30^{\circ},\) then which of the following describes the frequency and the angle of refraction in the chromium? (A) \(5 \times 10^{14} \mathrm{Hz} ; 9.6^{\circ}\) (B) \(5 \times 10^{14} \mathrm{Hz} ; 57^{\circ}\) (C) \(1.0 \times 10^{10} \mathrm{Hz} ; 9.6^{\circ}\) (D) \(1.0 \times 10^{10} \mathrm{Hz} ; 57^{\circ}\)

Short answer

Frequency is \( 5 \times 10^{14} \text{Hz} \) and angle of refraction in chromium is \( 9.6^{\circ} \). The correct option is (A).

Step-by-step solution

- Determine the frequency

The frequency of light does not change when traveling through different mediums. Therefore, the frequency in air, crystal, and chromium is the same. Given that the frequency in air is \( f = 5 \times 10^{14} \text{Hz} \), it remains \( 5 \times 10^{14} \text{Hz} \) in chromium.

- Use Snell's Law

Snell's Law relates the angles and indices of refraction of two media: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] where \( n_1 \) and \( n_2 \) are the indices of refraction, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction respectively.

- Calculate refraction angle at crystal

The light travels from air (\( n_1 = 1 \)) to crystal (\( n_2 = 2 \)). Using Snell's Law: \[ 1 \cdot \sin(30^{\circ}) = 2 \cdot \sin(\theta_{crystal}) \] With \( \sin(30^{\circ}) = \frac{1}{2} \): \[ \sin(\theta_{crystal}) = \frac{1}{2 \cdot 2} = \frac{1}{4} \] \( \theta_{crystal} = \sin^{-1}(\frac{1}{4}) \approx 14.48^{\circ} \)

- Calculate refraction angle in chromium

Next, the light travels from crystal (\( n_1 = 2 \)) to chromium (\( n_2 = 3 \)). Again applying Snell's Law: \[ 2 \cdot \sin(14.48^{\circ}) = 3 \cdot \sin(\theta_{chromium}) \] Using \( \sin(14.48^{\circ}) \approx 0.25 \):\[ 2 \cdot 0.25 = 3 \cdot \sin(\theta_{chromium}) \]\[ \sin(\theta_{chromium}) = \frac{2 \cdot 0.25}{3} = \frac{0.5}{3} \approx 0.1667 \] \( \theta_{chromium} = \sin^{-1}(0.1667) \approx 9.6^{\circ} \)

Key concepts

Snell's Law

The foundation of solving light refraction problems lies in understanding Snell's Law. Snell's Law helps us understand how light bends when it moves from one medium to another. The law is formulated as: Search Engine Optimization Utilizing the index of refraction of two mediums and the sin of the angles of incidence and refraction, you can find how light bends: Using this law, students can calculate the angle at which light enters a new medium. For example, going from air (n = 1) to crystal (n = 2), they would use Snell's Law to determine the angle of refraction in the crystal if the incident angle is 30 degrees.

Index of Refraction

The index of refraction indicates how much light bends when entering a material. It is a crucial part of Snell's Law and is represented by the letter 'n'. Different materials have different indices of refraction, which change how light travels through them.

For example:

• Air: n ≈ 1
• Water: n ≈ 1.33
• Crystal: n = 2
• Chromium: n = 3

When light enters a material with a higher index of refraction than the one it is coming from, it bends towards the normal (a line perpendicular to the surface). If it enters a material with a lower index, it bends away from the normal.
Knowing the indices can help you predict light behavior as it moves through different surfaces.

Angle of Incidence

The angle of incidence is the angle between the incoming ray of light and the normal at the point of contact with a surface. It plays a vital role in determining how light will bend when it enters a new medium.

When solving refraction problems, the angle of incidence is used in Snell's Law. For example, if a light ray hits crystal with an incidence angle of 30°, we use:
Where n1 in air is 1 and theta 1 is 30°.
Knowing the incidence angle and the refractive indices allows you to apply Snell's Law and find the refraction angle.

Frequency of Light

The frequency of light, represented by f, is the number of wave crests passing a point per second, measured in Hertz (Hz). Frequency does not change when light transitions between different mediums.

In the given exercise:

• Frequency in air: 5 × 10^14 Hz
• Frequency in crystal: 5 × 10^14 Hz
• Frequency in chromium: 5 × 10^14 Hz

This constancy is crucial because it means the energy of the light wave remains the same, even as its speed and wavelength change.
Understanding this property helps students differentiate between the changeable and unchangeable properties of light during refraction.