Problem 40 - Physics

Question

Violet light and red light travel through air and strike a block of plastic at the same angle of incidence. The angle of refraction is \(30.400^{\circ}\) for the violet light and \(31.200^{\circ}\) for the red light. The index of refraction for violet light in plastic is greater than that for red light by \(0.0400 .\) Delaying any rounding off of calculations until the very end, find the index of refraction for violet light in plastic.

Step-by-Step Solution

Answer

The index of refraction for violet light is approximately 1.5200.

Step 1: Understanding Snell's Law

Snell's Law relates the angle of incidence and the angle of refraction when light travels from one medium to another. It is given by:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]Where \( n_1 \) and \( n_2 \) are the indices of refraction of the first and second medium respectively, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction. In this problem, the values for \( n_2 \) will be different for violet and red light.

Step 2: Applying Snell's Law for Violet Light

For violet light, let the index of refraction for the plastic be \( n_v \). The air has an index of refraction \( n_1 = 1 \). The angle of refraction for violet light is given as \( \theta_v = 30.400^{\circ} \).Applying Snell's Law:\[ \sin(\theta_i) = n_v \sin(30.4^{\circ}) \]

Step 3: Applying Snell's Law for Red Light

For red light, let the index of refraction for the plastic be \( n_r \). The angle of refraction for red light is \( \theta_r = 31.200^{\circ} \), and we know \( n_v = n_r + 0.0400 \).Applying Snell's Law:\[ \sin(\theta_i) = n_r \sin(31.2^{\circ}) \]

Step 4: Expressing \( \sin(\theta_i) \) from Both Equations

From Step 2, we have:\[ \sin(\theta_i) = n_v \sin(30.4^{\circ}) \]From Step 3, we have:\[ \sin(\theta_i) = n_r \sin(31.2^{\circ}) \]Equating the two expressions for \( \sin(\theta_i) \), we can say:\[ n_v \sin(30.4^{\circ}) = n_r \sin(31.2^{\circ}) \]

Step 5: Substitute and Solve for \( n_r \)

We know from the problem statement that \( n_v = n_r + 0.0400 \). Substitute this into the equation from Step 4:\[ (n_r + 0.0400) \sin(30.4^{\circ}) = n_r \sin(31.2^{\circ}) \]Isolating \( n_r \), we get:\[ n_r (\sin(31.2^{\circ}) - \sin(30.4^{\circ})) = 0.0400 \sin(30.4^{\circ}) \]

Step 6: Calculate \( n_r \) Numerically

Calculate \( \sin(30.4^{\circ}) \) and \( \sin(31.2^{\circ}) \) using a calculator. Then solve for \( n_r \):\[ n_r \approx \frac{0.0400 \sin(30.4^{\circ})}{\sin(31.2^{\circ}) - \sin(30.4^{\circ})}\]

Step 7: Calculate \( n_v \) Using \( n_r \)

Once \( n_r \) is found from Step 6, use the relationship \( n_v = n_r + 0.0400 \) to find the index of refraction for violet light:\[ n_v \approx n_r + 0.0400 \]

Key Concepts

Index of Refraction

The index of refraction, denoted as \( n \), is a measure of how much a ray of light bends when it enters a new medium. Essentially, it's the ratio of the speed of light in a vacuum to its speed in the specified medium. In Snell's Law, the index of refraction plays a vital role in understanding how light transitions between different substances.

• In air, the index of refraction is approximately \( 1 \).
• For violet and red light, the indices of refraction in plastic differ slightly, as shown in the exercise.

Using the given information from the exercise, violet light has a slightly higher index of refraction compared to red light, precisely by 0.0400. This difference indicates that violet light bends more than red light when passing from air into a plastic medium.

Angle of Refraction

The angle of refraction is the measure of the angle made by the refracted ray with the normal to the surface at the point of refraction. This angle is determined by Snell's Law, where the initial and refracted angles are computed based on the indices of refraction of the involved mediums.

• The angle depends on the material light is entering.
• Each wavelength of light refracts at a slightly different angle due to its index of refraction.

In the given exercise:- Violet light refracts at an angle of \( 30.400^{\circ} \).- Red light, on the other hand, refracts at a slightly larger angle of \( 31.200^{\circ} \). The difference lies in their separate indices of refraction within the plastic medium.

Violet Light

Violet light, being at one end of the visible spectrum, has shorter wavelengths compared to other colors of visible light like red. This means violet light bends (refracts) differently when it travels through a medium like plastic.

• Considered the shortest wavelength in visible light.
• Has a greater index of refraction compared to longer wavelength light.

For the exercise, violet light has a refractive angle of \( 30.400^{\circ} \) and an index of refraction greater than red light by 0.0400. This greater index causes violet light to bend more sharply when it passes into the plastic medium, illustrating the concept that shorter wavelengths of light are refracted more than longer wavelengths.

Red Light

Red light, lying on the opposite end of the visible spectrum to violet, has longer wavelengths and consequently a lower index of refraction when it enters materials such as plastic.

• Has a longer wavelength than violet light.
• Consequently, has a smaller index of refraction in a given medium.

In the exercise, red light passes through the plastic at an angle of \( 31.200^{\circ} \). Its refracted angle is larger compared to violet light, indicating lesser bending due to its smaller index of refraction. This concept is key in understanding how different wavelengths of light behave distinctively upon entering a medium besides air.