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Chapter 25: Problem 66

Sonya is designing a diffraction experiment for her students. She has a laser that emits light of wavelength \(627 \mathrm{nm}\) and a grating with a distance of \(2.40 \times 10^{-3} \mathrm{mm}\) between slits. She hopes to shine the light through the grating and display a total of nine interference maxima on a screen. She finds that no matter how she arranges her setup, she can see only seven maxima. Assuming that the intensity of the light is not the problem, why can't Sonya display the \(m=4\) interference maxima on either side?

Short Answer

Expert verified
Answer: Sonya cannot observe an interference maxima with m=4 on either side because the maximum possible integer value of m that can be observed in this setup is 3. This is because higher-order maxima (m=4) would go beyond the maximum possible angle for interference maxima according to the diffraction grating interference formula.

Step by step solution

01

Understand the diffraction grating interference formula

The formula used to analyze diffraction grating interference is given by: \(d \cdot \sin{\theta} = m \cdot \lambda\) Where \(d\) is the distance between slits, \(\theta\) is the angle of the light's path with respect to the grating's normal, \(m\) is the order of the interference maxima, and \(\lambda\) is the wavelength of the light.
02

Identify the given information

We know the following information: - Wavelength of light (\(\lambda\)) = \(627 \times 10^{-9} \mathrm{m}\) (converting nm to m) - Distance between slits in the grating (\(d\)) = \(2.40 \times 10^{-3} \mathrm{mm} = 2.40 \times 10^{-6} \mathrm{m}\) (converting mm to m) - We are asked to verify if the maximum possible \(m\) is indeed \(3\) and not \(4\).
03

Find the maximum possible value of \(\sin{\theta}\)

The maximum value of \(\sin{\theta}\) is 1, which occurs when \(\theta\) approaches \(90^{\circ}\). Let's use this observation to find the maximum value of \(m\).
04

Calculate the value of \(m_{max}\)

Using the formula, we can substitute the values of \(\lambda\), \(d\), and \(\sin{\theta_{max}}=1\) to find the maximum value of \(m\). Let's call this value \(m_{max}\): \(d \cdot 1 = m_{max} \cdot \lambda\) Now, we can find the value of \(m_{max}\): \(m_{max} = \frac{d}{\lambda} = \frac{2.40 \times 10^{-6} \mathrm{m}}{627 \times 10^{-9} \mathrm{m}} \approx 3.83\)
05

Interpret the result

Since \(m_{max} \approx 3.83\), it means that the maximum possible integer value of \(m\) that Sonya can observe is \(3\) as higher-order maxima (\(m=4\)) would go beyond the maximum possible angle for interference maxima. Therefore, it is not possible for Sonya to observe \(m=4\) interference maxima on either side of her setup.

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Most popular questions from this chapter

A thin layer of an oil \((n=1.60)\) floats on top of water \((n=1.33) .\) One portion of this film appears green \((\lambda=510 \mathrm{nm})\) in reflected light. How thick is this portion of the film? Give the three smallest possibilities.
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?
Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]
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?
Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)
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