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Chapter 25: Problem 27

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?

Short Answer

Expert verified
Based on the given information that the distances between the primary red and violet colors on the screen is 7.00 mm, and the distance from the slide to the screen is 3.40 meters, we calculated the separation between the two slits using the double-slit interference formula. The separation between the two slits is approximately 146 micrometers.

Step by step solution

01

Understand the double-slit interference formula

The double-slit interference formula is given by: \(\sin \theta = \frac{m \lambda}{d}\) where \(\theta\) is the angle between the central maximum and the m-th order maximum on the screen, \(m\) is the order of the interference pattern, \(\lambda\) is the wavelength of light, and \(d\) is the separation between the slits.
02

Calculate the angles for red and violet light

In this problem, we are given the distances between the red and violet light on the screen and the distance from the slide to the screen. We can use the small angle approximation (\(\sin \theta \approx \tan \theta\)) to calculate the angles for red and violet light: \(\tan \theta_{red} = \frac{distance_{red}}{distance_{screen}}\) \(\tan \theta_{violet} = \frac{distance_{violet}}{distance_{screen}}\) \(\tan \theta_{red} - \tan \theta_{violet} = \frac{7.00\,\mathrm{mm}}{3.40\,\mathrm{m}}\)
03

Use the double-slit interference formula to find the separation between slits

Since we are dealing with the first-order maximum (\(m=1\)), we can substitute the wavelengths and the calculated angle difference into the double-slit interference formula: \(\frac{\sin \theta_{red} - \sin \theta_{violet}}{1} = \frac{\lambda_{red} - \lambda_{violet}}{d}\) Using the small angle approximation, we get: \(\frac{\tan \theta_{red} - \tan \theta_{violet}}{1} = \frac{\lambda_{red} - \lambda_{violet}}{d}\) Plugging in the known values, we have: \(\frac{7.00\,\mathrm{mm}}{3.40\,\mathrm{m}} = \frac{700\,\mathrm{nm} - 400\,\mathrm{nm}}{d}\) Now we can solve for the separation between the slits, \(d\).
04

Solve for the slit separation, \(d\)

Rearrange the equation to find \(d\): \(d = \frac{(700\,\mathrm{nm} - 400\,\mathrm{nm})(3.40\,\mathrm{m})}{7.00\,\mathrm{mm}}\) Perform the calculation: \(d = \frac{300\,\mathrm{nm}(3.40\,\mathrm{m})}{7.00\,\mathrm{mm}}\) \(d = 0.000146\,\mathrm{m} = 146\,\mathrm{\mu m}\)
05

Conclusion

The separation between the two slits is approximately \(146\,\mathrm{\mu m}\).

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Most popular questions from this chapter

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?
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 \(\boldsymbol{\theta} .]\)
Diffraction by a single Slit The central bright fringe in a single-slit diffraction pattern from light of wavelength 476 nm is \(2.0 \mathrm{cm}\) wide on a screen that is $1.05 \mathrm{m}$ from the slit. (a) How wide is the slit? (b) How wide are the first two bright fringes on either side of the central bright fringe? (Define the width of a bright fringe as the linear distance from minimum to minimum.)
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?
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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