/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 In a double-slit interference ex... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a double-slit interference experiment, the wavelength is \(475 \mathrm{nm}\), the slit separation is \(0.120 \mathrm{mm},\) and the screen is $36.8 \mathrm{cm}$ away from the slits. What is the linear distance between adjacent maxima on the screen? [Hint: Assume the small-angle approximation is justified and then check the validity of your assumption once you know the value of the separation between adjacent maxima.] (tutorial: double slit 1 ).

Short Answer

Expert verified
Answer: The linear distance between adjacent maxima on the screen is approximately 1.457 mm.

Step by step solution

01

Convert all units to meters

To have consistent units in the calculations, we need to convert all the given values to meters. We have, \(\lambda = 475 \thinspace nm = 475 \times 10^{-9} \thinspace m\), \(d = 0.120 \thinspace mm = 0.120 \times 10^{-3} \thinspace m\), and \(L = 36.8 \thinspace cm = 36.8 \times 10^{-2} \thinspace m\).
02

Use the small-angle approximation formula for the angular separation

Under the small-angle approximation, the angular separation between adjacent maxima (\(θ\)) for a double-slit interference experiment can be calculated using the formula: \(θ = \frac{m\lambda}{d}\), where \(m\) is the order of the maxima (e.g., \(1\) for the first maxima, \(2\) for the second maxima, etc.). Let's consider the distance between the first maxima on either side of the central maxima (i.e., \(m = 1\)). This will give us the distance between adjacent maxima. \(θ = \frac{\lambda}{d} = \frac{475 \times 10^{-9} \thinspace m}{0.120 \times 10^{-3} \thinspace m} = 3.958 \times 10^{-3}\) radians
03

Calculate the linear distance

Now, we need to convert the angular separation (\(θ\)) between adjacent maxima into a linear distance (\(y\)) on the screen. We use the following formula for that: \(y = L\cdot tan(θ) \approx L\cdot θ \thinspace\) (assuming the small-angle approximation) Now, plugging the values, we get: \(y = 36.8 \times 10^{-2} \thinspace m \cdot 3.958 \times 10^{-3}\) radians \(y = 1.457 \times 10^{-3} \thinspace m = 1.457 \thinspace mm\) So, the linear distance between adjacent maxima on the screen is approximately \(1.457 \thinspace mm\).
04

Check the validity of the small-angle approximation

To confirm that the small-angle approximation is valid, we should check if \(tan(θ)\) is approximately equal to \(θ\). If it is, then our answer is accurate. Let's check that: \(tan(3.958 \times 10^{-3}) \approx 3.958 \times 10^{-3}\) Using a calculator, we find that \(tan(3.958 \times 10^{-3}) \approx 3.959 \times 10^{-3}\), which is very close to \(3.958 \times 10^{-3}\). This means that the small-angle approximation is valid and our calculated linear distance between adjacent maxima is accurate.

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 91Ó°ÊÓ!

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

Two slits separated by \(20.0 \mu \mathrm{m}\) are illuminated by light of wavelength \(0.50 \mu \mathrm{m} .\) If the screen is \(8.0 \mathrm{m}\) from the slits, what is the distance between the \(m=0\) and \(m=1\) bright fringes?
A red line (wavelength \(630 \mathrm{nm}\) ) in the third order overlaps with a blue line in the fourth order for a particular grating. What is the wavelength of the blue line?
Suppose a transparent vessel \(30.0 \mathrm{cm}\) long is placed in one arm of a Michelson interferometer, as in Example 25.2. The vessel initially contains air at \(0^{\circ} \mathrm{C}\) and 1.00 atm. With light of vacuum wavelength \(633 \mathrm{nm}\), the mirrors are arranged so that a bright spot appears at the center of the screen. As air is slowly pumped out of the vessel, one of the mirrors is gradually moved to keep the center region of the screen bright. The distance the mirror moves is measured to determine the value of the index of refraction of air, \(n .\) Assume that, outside of the vessel, the light travels through vacuum. Calculate the distance that the mirror would be moved as the container is emptied of air.
If you shine a laser (wavelength \(0.60 \mu \mathrm{m}\) ) with a small aperture at the Moon, diffraction makes the beam spread out and the spot on the Moon is large. Making the aperture smaller only makes the spot on the Moon larger. On the other hand, shining a wide searchlight at the Moon can't make a tiny spot- the spot on the Moon is at least as wide as the searchlight. What is the radius of the smallest possible spot you can make on the Moon by shining a light from Earth? Assume the light is perfectly parallel before passing through a circular aperture.
A grating has 5000.0 slits/cm. How many orders of violet light of wavelength \(412 \mathrm{nm}\) can be observed with this grating? (tutorial: grating).
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.