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Chapter 4: Problem 118

Blue light of wavelength 450 nm falls on a slit of width \(0.25 \mathrm{mm}\). A converging lens of focal length \(20 \mathrm{cm}\) is placed behind the slit and focuses the diffraction pattern on a screen. (a) How far is the screen from the lens? (b) What is the distance between the first and the third minima of the diffraction pattern?

Short Answer

Expert verified
The screen is located \(20\, \mathrm{cm}\) away from the lens. The distance between the first and the third minima of the diffraction pattern is \(y_{13} = y_3 - y_1\). Calculate \(y_1\) and \(y_3\) using \(\tan(\theta_1) = \frac{y_1}{v}\) and \(\tan(\theta_3) = \frac{y_3}{v}\) with the angle values from step 2 and the screen distance from step 3.

Step by step solution

01

Recall the Fraunhofer diffraction formula for single slit minima

Recall the formula for the positions of the minima in a single-slit Fraunhofer diffraction pattern: \(a\sin\theta = m\lambda\), where: \(a\) is the slit width, \(\theta\) is the angle of the minimum from the central maximum, \(m\) is the order of the minimum (1 for the first minimum, 2 for the second minimum, etc.), and \(\lambda\) is the wavelength of light.
02

Calculate the angle of the first and the third minima

From the formula mentioned above, we will compute the angle for the first and the third minima. We have the values of \(a = 0.25\, \mathrm{mm}\), and \(\lambda = 450\, \mathrm{nm}\). For the first minimum (m = 1): \(\sin\theta_1 = \frac{1 \times 450\,\mathrm{nm}}{0.25\,\mathrm{mm}}\), For the third minimum (m = 3): \(\sin\theta_3 = \frac{3 \times 450\,\mathrm{nm}}{0.25\,\mathrm{mm}}\).
03

Use the thin lens equation to find the screen distance

The thin lens equation is given by: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), where: \(f\) is the focal length of the lens, \(u\) is the distance between the object and the lens, and \(v\) is the distance between the image and the lens. We are given that the focal length of the converging lens is \(20\, \mathrm{cm}\). Since the Fraunhofer diffraction pattern is focused at infinity, we have \(u = \infty\). Thus, \(\frac{1}{20\, \mathrm{cm}} = 0 + \frac{1}{v}\), From this equation, we can solve for \(v\), the distance between the lens and the screen.
04

Calculate the distance between the first and the third minima on the screen

Using the small-angle approximation \(\sin(\theta) \approx \tan(\theta)\), we can write: \(\tan(\theta_1) = \frac{y_1}{v}\), \(\tan(\theta_3) = \frac{y_3}{v}\), where: \(y_1\) is the distance between the central maximum and the first minimum on the screen, \(y_3\) is the distance between the central maximum and the third minimum on the screen, and \(v\) is the distance between the lens and the screen. Now, we can find \(y_1\) and \(y_3\) using the angle values we obtained in step 2 and the screen distance obtained in step 3. Finally, subtract the distances to find the distance between the first and the third minima on the screen: \(y_{13} = y_3 - y_1\).

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Most popular questions from this chapter

How far apart must two objects be on the moon to be distinguishable by eye if only the diffraction effects of the eye's pupil limit the resolution? Assume \(550 \mathrm{nm}\) for the wavelength of light, the pupil diameter \(5.0 \mathrm{mm}\), and \(400,000 \mathrm{km}\) for the distance to the moon.

How narrow is a slit that produces a diffraction pattem on a screen \(1.8 \mathrm{m}\) away whose central peak is \(1.0 \mathrm{m}\) wide? Assume \(\lambda=589 \mathrm{nm}\)

When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh's criterion. Take the pupil of the eye to be \(3.0 \mathrm{mm}\) and the distance from the paper to the eye of \(35 \mathrm{cm}\); find the minimum separation of two dots such that they cannot be resolved. How many dots per inch (dpi) does this correspond to?

(a) Assume that the maxima are halfway between the minima of a single-slit diffraction pattern. The use the diameter and circumference of the phasor diagram, as described in Intensity in single-Slit Diffraction, to determine the intensities of the third and fourth maxima in terms of the intensity of the central maximum. (b) Do the same calculation, using Equation 4.4.

A double slit produces a diffraction pattern that is a combination of single- and double-slit interference. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single-slit pattern falls on the fifth maximum of the double-slit pattern. (This will greatly reduce the intensity of the fifth maximum.)
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