Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A glass plate 2.50 mm thick, with an index of refraction of 1.40, is placed between a point source of light with wavelength 540 nm (in vacuum) and a screen. The distance from source to screen is 1.80 cm. How many wavelengths are there between the source and the screen?

Short Answer

Expert verified
9,349 wavelengths are between the source and the screen.

Step by step solution

01

Convert Given Units

Convert the thickness of the glass plate from millimeters to centimeters. We know that 1 mm = 0.1 cm, so 2.50 mm = 0.25 cm.
02

Calculate the Wavelength in Glass

In a medium other than vacuum, the wavelength changes. Use the formula \( \lambda_m = \frac{\lambda_0}{n} \), where \( \lambda_0 = 540 \) nm and \( n = 1.40 \). Thus, \( \lambda_m = \frac{540 \text{ nm}}{1.40} \approx 385.7 \text{ nm} \). Convert this to centimeters: \( 385.7 \text{ nm} \approx 3.857 \times 10^{-5} \text{ cm} \).
03

Determine the Number of Wavelengths in Glass

Using the wavelength in the glass, calculate how many wavelengths fit in the 0.25 cm thickness of the glass. \( N_{glass} = \frac{0.25 \text{ cm}}{3.857 \times 10^{-5} \text{ cm}} \approx 6479 \text{ wavelengths} \).
04

Determine the Number of Wavelengths in Air

Calculate the number of wavelengths corresponding to the remaining distance (1.80 cm - 0.25 cm = 1.55 cm) using \( \lambda_0 = 540 \) nm, which equals \( 5.40 \times 10^{-5} \) cm. Therefore, \( N_{air} = \frac{1.55 \text{ cm}}{5.40 \times 10^{-5} \text{ cm}} \approx 2870 \text{ wavelengths}. \)
05

Total Number of Wavelengths

Add the wavelengths from the path through the glass and the path through the air to find the total number of wavelengths: \( N_{total} = N_{glass} + N_{air} = 6479 + 2870 = 9349 \text{ wavelengths}. \)
06

Conclusion: Result Interpretation

The total number of wavelengths between the source and the screen is the sum of wavelengths in the glass and air segments. This gives us the comprehensive optical distance in terms of the original light wavelength.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Index of Refraction
The index of refraction, often denoted as \( n \), is a measure of how much light slows down in a medium compared to its speed in a vacuum. This concept is crucial in wave optics because it influences how light behaves as it travels through different materials. Here's how it works:
  • In a vacuum, light travels at its maximum speed, approximately \( 3 \times 10^8 \) meters per second.
  • When light enters a medium, like glass or water, it slows down, resulting in a lower speed.
  • The index of refraction is calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium: \( n = \frac{c}{v} \).
For example, if the index of refraction of glass is 1.40, this means light travels 1.40 times slower in glass than in a vacuum. This slowing effect alters not only the speed but also the path and, crucially, the wavelength of light in the medium.
Wavelength in Medium
Wavelength refers to the distance between consecutive crests or troughs of a wave. In optics, the wavelength of light is a fundamental property. However, it is not always constant; it changes when light moves through different materials.The wavelength in a medium is given by \( \lambda_m = \frac{\lambda_0}{n} \), where \( \lambda_0 \) is the wavelength in a vacuum, and \( n \) is the index of refraction.
  • If the vacuum wavelength is 540 nm and the index of refraction of the glass is 1.40, the wavelength in the glass becomes approximately 385.7 nm.
  • This change is due to the fact that the speed of light is reduced in the medium, causing the wave “compression.”
This altered wavelength is essential for calculations involving phase and optical path length, allowing us to determine how the wave propagates through the medium.
Optical Path Length
Optical path length (OPL) is a concept that helps us understand how light travels in different media. It combines the physical distance the light travels with the medium's effect on light. The formula for optical path length is:\[OPL = n \, \times \text{ physical distance}.\]
  • For example, if light travels through a 0.25 cm thick glass plate with an index of refraction of 1.40, the OPL becomes 0.35 cm.
  • Similarly, light traveling through air over a distance of 1.55 cm maintains almost the same OPL, assuming air's refractive index is close to 1.
The optical path length is foundational for calculating the phase shift in light and determining how many wavelengths fit between two points, which is fundamental in wave optics for understanding light behavior and interference.
Light Transmission
Light transmission describes how light waves pass through a material. Whether reflecting, refracting, or being absorbed, each material affects light differently. In contexts like lenses and optical fibers, understanding light transmission becomes crucial:
  • Materials with higher close-to-unity transmittance allow more light to pass through with minimal loss.
  • The glass plate, with its index of refraction, will affect the light's transmission by altering its speed and wavelength, creating potential for reflection or absorption.
For applications relying on accurate light passage, adjusting for the index of refraction ensures precision. By calculating light behavior through these media, engineers and scientists can enhance device design, improving everything from eyeglasses to advanced communication systems, fostering better technological integration into daily life.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A light beam travels at \(1.94 \times 10^8\) m/s in quartz. The wavelength of the light in quartz is 355 nm. (a) What is the index of refraction of quartz at this wavelength? (b) If this same light travels through air, what is its wavelength there?

A ray of light is traveling in a glass cube that is totally immersed in water. You find that if the ray is incident on the glass-water interface at an angle to the normal larger than 48.7\(^\circ\), no light is refracted into the water. What is the refractive index of the glass?

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at 23.0\(^\circ\) and 62.0\(^\circ\), respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 55.0 \(\mathrm {W/cm}^2\) after it passes through the stack. If the incident intensity is kept constant but the second polarizer is removed, what is the intensity of the light after it has passed through the stack?

The refractive index of a certain glass is 1.66. For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?

The indexes of refraction for violet light \((\lambda = 400 \, \mathrm{nm})\) and red light \((\lambda = 700 \, \mathrm{nm})\) in diamond are 2.46 and 2.41, respectively. A ray of light traveling through air strikes the diamond surface at an angle of 53.5\(^\circ\) to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free