Red light of 650 nm can be seen in three orders in a particular grating. About how many slits per centimeter does this grating have?

Short Answer

Expert verified
Answer: The diffraction grating has approximately 15,385 slits/cm.

Step by step solution

01

Understand the diffraction grating formula

The formula for the diffraction grating is given by: d * sin(θ) = m * λ where, d = distance between slits (or the grating constant), θ = angle at which maximum intensity occurs, m = order, λ = wavelength of the light. We can rewrite the formula to find the d value: d = (m * λ) / sin(θ) We are given λ = 650 nm and we need to find d.
02

Calculate the angle θ for each order

Since we are given that the red light can be seen in three orders, we will calculate the angle θ for each order (m = 1, 2, 3). For m = 1, the angle θ will be such that sin(θ) = λ / d. Since we have no information about θ, we can assume the value of θ to be small. In that case sin(θ) ≈ tan(θ) ≈ θ, and we can rewrite the formula as: d ≈ (1 * λ) / θ We can follow the same approximation for m = 2 and m = 3: For m = 2: d ≈ (2 * λ) / θ For m = 3: d ≈ (3 * λ) / θ
03

Find the grating constant (d) and number of slits per centimeter

Now, we will use the three equations to eliminate the θ variable and find the grating constant (d). Divide the equation for m = 2 by the equation for m = 1: (2 * λ) / θ ≈ 2 * (λ / θ) which simplifies to d ≈ 2 * d Similarly, For m = 3, divide the equation for m = 3 by the equation for m = 1: (3 * λ) / θ ≈ 3 * (λ / θ) which simplifies to d ≈ 3 * d Now, we have two equations for d: For m = 2: d ≈ 2 * d For m = 3: d ≈ 3 * d Solving the two equations, we find that d must be such that it is an integral multiple of itself, mainly d = n * 650 nm, where n is a natural number. If we consider the lowest possible value for n (n = 1), we will have the minimum distance between the slits (d_min). d_min = 1 * 650 nm = 650 nm
04

Calculate the number of slits per centimeter

The number of slits per centimeter can be found by converting d_min from nanometers to centimeters and then finding the reciprocal: d_min = 650 nm * (1 cm / 1e7 nm) = 6.5e-5 cm Number of slits per centimeter = 1 / d_min = 1 / 6.5e-5 cm = 15385 slits/cm The grating has approximately 15,385 slits per centimeter.

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 mica sheet \(1.00 \mu \mathrm{m}\) thick is suspended in air. In reflected light, there are gaps in the visible spectrum at \(450,525,\) and $630 \mathrm{nm} .$ Calculate the index of refraction of the mica sheet.
Thin Films At a science museum, Marlow looks down into a display case and sees two pieces of very flat glass lying on top of each other with light and dark regions on the glass. The exhibit states that monochromatic light with a wavelength of \(550 \mathrm{nm}\) is incident on the glass plates and that the plates are sitting in air. The glass has an index of refraction of \(1.51 .\) (a) What is the minimum distance between the two glass plates for one of the dark regions? (b) What is the minimum distance between the two glass plates for one of the light regions? (c) What is the next largest distance between the plates for a dark region? [Hint: Do not worry about the thickness of the glass plates; the thin film is the air between the plates.]
A camera lens \((n=1.50)\) is coated with a thin film of magnesium fluoride \((n=1.38)\) of thickness \(90.0 \mathrm{nm}\) What wavelength in the visible spectrum is most strongly transmitted through the film?
A pinhole camera doesn't have a lens; a small circular hole lets light into the camera, which then exposes the film. For the sharpest image, light from a distant point source makes as small a spot on the film as possible. What is the optimum size of the hole for a camera in which the film is $16.0 \mathrm{cm}$ from the pinhole? A hole smaller than the optimum makes a larger spot since it diffracts the light more. A larger hole also makes a larger spot because the spot cannot be smaller than the hole itself (think in terms of geometrical optics). Let the wavelength be \(560 \mathrm{nm}\).
You are given a slide with two slits cut into it and asked how far apart the slits are. You shine white light on the slide and notice the first-order color spectrum that is created on a screen \(3.40 \mathrm{m}\) away. On the screen, the red light with a wavelength of 700 nm is separated from the violet light with a wavelength of 400 nm by 7.00 mm. What is the separation of the two slits?
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