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Chapter 26: Problem 39

Red light \((n=1.520)\) and violet light \((n=1.538)\) traveling in air are incident on a slab of crown glass. Both colors enter the glass at the same angle of refraction. The red light has an angle of incidence of \(30.00^{\circ} .\) What is the angle of incidence of the violet light?

Short Answer

Expert verified
The angle of incidence of the violet light is approximately \(30.39^{\circ}\).

Step by step solution

01

Identify Given Data

We have two types of light, red with a refractive index of \(n_{red} = 1.520\) and violet with a refractive index of \(n_{violet} = 1.538\). The angle of incidence of red light is \(30.00^{\circ}\). We need to find the angle of incidence of violet light.
02

Use Snell's Law for Red Light

According to Snell's Law, \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). For red light, where \( n_1 = 1 \) (refractive index of air), we have:\[1 \times \sin(30.00^{\circ}) = 1.520 \times \sin(\theta_{red})\]We solve for \( \sin(\theta_{red}) \).
03

Calculate Sin of Red's Refraction Angle

Calculate \( \sin(\theta_{red}) \) using:\[\sin(\theta_{red}) = \frac{\sin(30.00^{\circ})}{1.520} \approx \frac{0.5}{1.520} \approx 0.3289\]And find \( \theta_{red} \) using \( \arcsin(0.3289) \).
04

Use Snell's Law for Violet Light

Use Snell's Law again, assuming both colors have the same \( \theta_{glass} \), the angle of refraction in glass:\[1 \times \sin(\theta_{violet}) = 1.538 \times \sin(\theta_{red})\]Solve for \( \sin(\theta_{violet}) \).
05

Calculate Angle of Incidence for Violet Light

From the previous step, we use the already calculated \( \sin(\theta_{red}) \):\[\sin(\theta_{violet}) = \frac{1.538 \times 0.3289}{1} \approx 0.5059\]Now, solve for \( \theta_{violet} \) using \( \arcsin(0.5059) \).
06

Final Calculation for Violet

Now calculate \( \theta_{violet} \) which gives the angle of incidence:\[\theta_{violet} \approx \arcsin(0.5059) \approx 30.39^{\circ}\]This is the angle of incidence for violet light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index is a measure that indicates how much light bends when it enters a medium. It is a crucial component in understanding light behavior as it moves through different substances. Here's why it matters:
  • It helps us determine how much a ray of light will bend when it transitions from one medium to another.
  • A higher refractive index means light will bend more significantly.
  • It is calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium.
In our exercise, the refractive index for red light in crown glass is 1.520 and for violet light is 1.538. These values indicate that violet light will bend slightly more than red light upon entering crown glass from air.
Angle of Incidence
When a ray of light strikes a surface, it forms an angle with a line perpendicular to that surface, known as the normal line. This angle is called the angle of incidence. Here's what you need to know:
  • The angle of incidence is measured from the normal to the incoming ray of light.
  • It is crucial in determining how much the ray will bend on entering a different medium according to Snell's Law.
In the scenario addressed, the red light has an angle of incidence of 30.00掳 as it hits the crown glass. To find the angle of incidence for violet light, we used Snell's Law, considering that both colors share the same angle of refraction.
Angle of Refraction
The angle of refraction is the angle between the refracted ray and the normal line inside the new medium. Understanding this angle is essential for analyzing how light travels through different substances.
  • It changes depending on the refractive indices of the substances light moves between.
  • Snell's Law relates the angle of incidence and the angle of refraction through the indices of both mediums involved.
In our exercise, both red and violet lights share the same angle of refraction within the crown glass. By applying Snell's Law, we calculated this angle based on the red light details to then infer the angle needed for violet light.
Crown Glass
Crown glass is a type of optical glass noted for its durability and optical clarity. Here鈥檚 why it is important in optics:
  • It has a relatively low dispersion, meaning it minimizes the separation of light into its component colors.
  • With moderate refractive indices like 1.520 for red and 1.538 for violet light, it allows for precise control over light bending.
In our problem, crown glass serves as the medium into which both red and violet lights refract. Its characteristics help explain why light bends differently based on color, illustrating key principles of optics.

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Most popular questions from this chapter

To focus a camera on objects at different distances, the converging lens is moved toward or away from the film, so a sharp im age always falls on the film. A camera with a telephoto lens \((f=200.0 \mathrm{~mm})\) is to be focused on an object located first at a distance of \(3.5 \mathrm{~m}\) and then at \(50.0 \mathrm{~m}\). Over what distance must the lens be movable?

Two identical diverging lenses are separated by \(16 \mathrm{~cm} .\) The focal length of each lens is \(-8.0 \mathrm{~cm} .\) An object is located \(4.0 \mathrm{~cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

Two converging lenses are separated by \(24.00 \mathrm{~cm} .\) The focal length of each lens is 12.00 \(\mathrm{cm}\). An object is placed \(36.00 \mathrm{~cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

An amateur astronomer decides to build a telescope from a discarded pair of eyeglasses. One of the lenses has a refractive power of 11 diopters, and the other has a refractive power of 1.3 diopters. (a) Which lens should be the objective? (b) How far apart should the lenses be separated? (c) What is the angular magnification of the telescope?

A microscope for viewing blood cells has an objective with a focal length of 0.50 \(\mathrm{cm}\) and an eyepiece with a focal length of \(2.5 \mathrm{~cm} .\) The distance between the objective and eyepiece is \(14.0 \mathrm{~cm} .\) If a blood cell subtends an angle of \(2.1 \times 10^{-5}\) rad when viewed with the naked eye at a near point of \(25.0 \mathrm{~cm}\), what angle (magnitude only) does it subtend when viewed through the microscope?
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