Gratings A grating has exactly 8000 slits uniformly spaced over \(2.54 \mathrm{cm}\) and is illuminated by light from a mercury vapor discharge lamp. What is the expected angle for the third-order maximum of the green line $(\lambda=546 \mathrm{nm}) ?$

Short Answer

Expert verified
The expected angle for the third-order maximum of the green line (λ = 546 nm) is approximately 31.4°.

Step by step solution

01

Identify the grating equation and the given variables

In this exercise, we will use the grating equation:$$m\lambda = d\sin\theta$$Where \(m\) is the order of the maximum, \(\lambda\) is the wavelength of light, \(d\) is the slit separation, and \(\theta\) is the angle we need to find. We are given: - The green line wavelength: \(\lambda = 546\mathrm{nm} = 5.46 \times 10^{-7}\mathrm{m}\) (converted to meters) - The total distance of 8000 slits: \(2.54\mathrm{cm} = 2.54 \times 10^{-2}\mathrm{m}\) (converted to meters) - The order of the maximum we are looking for: \(m = 3\)
02

Calculate the slit separation, d

To find the slit separation, we can simply divide the total distance by the number of slits:$$d = \frac{2.54 \times 10^{-2}\mathrm{m}}{8000}$$Calculating this value gives us:$$d = 3.175 \times 10^{-6}\mathrm{m}$$
03

Solve the grating equation for the angle, θ

Now that we have the slit separation, we can plug in the values into the grating equation and solve for the angle:$$3(5.46 \times 10^{-7}\mathrm{m}) = (3.175 \times 10^{-6}\mathrm{m})\sin\theta$$Divide by \(3.175 \times 10^{-6}\mathrm{m}\) on both sides to isolate \(\sin\theta\):$$\sin\theta = \frac{3(5.46 \times 10^{-7}\mathrm{m})}{3.175 \times 10^{-6}\mathrm{m}}$$Calculate the value of \(\sin\theta\):$$\sin\theta \approx 0.5164$$
04

Find the angle θ

Finally, to find the angle, we take the inverse sine of the calculated value:$$\theta = \sin^{-1}(0.5164)$$Calculating this value gives us:$$\theta \approx 31.4°$$ The expected angle for the third-order maximum of the green line \((\lambda=546\mathrm{nm})\) is approximately \(31.4°\).

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!

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 thin soap film \((n=1.35)\) is suspended in air. The spectrum of light reflected from the film is missing two visible wavelengths of $500.0 \mathrm{nm}\( and \)600.0 \mathrm{nm},$ with no missing wavelengths between the two. (a) What is the thickness of the soap film? (b) Are there any other visible wavelengths missing from the reflected light? If so, what are they? (c) What wavelengths of light are strongest in the transmitted light?
The radio telescope at Arecibo, Puerto Rico, has a reflecting spherical bowl of \(305 \mathrm{m}(1000 \mathrm{ft})\) diameter. Radio signals can be received and emitted at various frequencies with appropriate antennae at the focal point of the reflecting bowl. At a frequency of \(300 \mathrm{MHz}\), what is the angle between two stars that can barely be resolved? (Tutorial:radio telescope).
Parallel light of wavelength \(\lambda\) strikes a slit of width \(a\) at normal incidence. The light is viewed on a screen that is \(1.0 \mathrm{m}\) past the slits. In each case that follows, sketch the intensity on the screen as a function of \(x\), the distance from the center of the screen, for $0 \leq x \leq 10 \mathrm{cm}$ (a) \(\lambda=10 a\). (b) \(10 \lambda=a,\) (c) \(30 \lambda=a.\)
A thin film of oil \((n=1.50)\) of thickness \(0.40 \mu \mathrm{m}\) is spread over a puddle of water \((n=1.33) .\) For which wavelength in the visible spectrum do you expect constructive interference for reflection at normal incidence?
Show that the interference fringes in a double-slit experiment are equally spaced on a distant screen near the center of the interference pattern. [Hint: Use the small angle approximation for \(\boldsymbol{\theta} .]\)
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