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Chapter 4: Problem 80

Microwaves of wavelength 10.0 mm fall normally on a metal plate that contains a slit \(25 \mathrm{mm}\) wide. (a) Where are the first minima of the diffraction pattern? (b) Would there be minima if the wavelength were \(30.0 \mathrm{mm}\) ?

Short Answer

Expert verified
(a) The first minima of the diffraction pattern occurs at angles of approximately \(\pm 23.6^\circ\). (b) There would not be a minima if the wavelength were \(30.0\,\text{mm}\).

Step by step solution

01

Write down the given information and the single-slit diffraction formula

Given the wavelength (\(\lambda\)) of microwaves: \(10\,\text{mm}\), and the slit width (\(a\)): \(25\,\text{mm}\). The single-slit diffraction formula is: \[ \sin \theta = \frac{m\lambda}{a} \] where \(\theta\) is the angle between the central maximum and the m-th order minimum, and \(m\) is the order of the minima.
02

Calculate the angle for the first minima

In this problem, we need to find the angular positions of the first minima, so we will use \(m = 1\). Plugging the values into the formula, we have: \[ \sin \theta_1 = \frac{1 \times 10}{25} \] Now, calculate the value of \(\sin \theta_1\): \[ \sin \theta_1 = \frac{10}{25} = 0.4 \]
03

Calculate the angle for the first minima

To find the angle for the first minima (\(\theta_1\)), take the inverse sine of the obtained value: \[ \theta_1 = \sin^{-1}(0.4) \approx 23.6^\circ \]
04

Determine if there are minima with a wavelength of 30.0 mm

Now, we need to check if there would be minima for a wavelength of \(30.0\,\text{mm}\). Plugging the new wavelength value \(\lambda = 30.0\,\text{mm}\) and \(m = 1\) into the formula, we get: \[ \sin \theta'_1 = \frac{1 \times 30}{25} \] Calculate the value of \(\sin \theta'_1\): \[ \sin \theta'_1 = \frac{30}{25} = 1.2 \] Since \(\sin \theta'_1 > 1\), it's not possible to have a minima for a wavelength of \(30.0\,\text{mm}\).
05

Present the results

(a) The first minima of the diffraction pattern occurs at angles of approximately \(\pm 23.6^\circ\). (b) There would not be a minima if the wavelength were \(30.0\,\text{mm}\).

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Most popular questions from this chapter

(a) How wide is a single slit that produces its first minimum for 633 -nm light at an angle of \(28.0^{\circ}\) ? (b) At what angle will the second minimum be?

(a) What is the minimum angular spread of a 633 -nm wavelength He-Ne laser beam that is originally \(1.00 \mathrm{mm}\) in diameter? (b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be? (c) How big a spot would be illuminated on the moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round- trip time and, hence, distance.)

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X-rays of wavelength 0.103 nm reflects off a crystal and a second-order maximum is recorded at a Bragg angle of \(25.5^{\circ}\). What is the spacing between the scattering planes in this crystal?

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