Chapter 33: Problem 43
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: \( n = \frac{c}{v} \).
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.”
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.
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.