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Chapter 25: Problem 72

Parallel light of wavelength \(\lambda\) strikes a slit of width \(a\) at normal incidence. The light is viewed on a screen that is \(1.0 \mathrm{m}\) past the slits. In each case that follows, sketch the intensity on the screen as a function of \(x\), the distance from the center of the screen, for $0 \leq x \leq 10 \mathrm{cm}$ (a) \(\lambda=10 a\). (b) \(10 \lambda=a,\) (c) \(30 \lambda=a.\)

Short Answer

Expert verified
Short Answer: To analyze the intensity pattern of parallel light hitting a single slit of different widths, we used the intensity formula \(I(\theta) = I_0 \left[ \frac{\sin(\beta)}{\beta} \right]^2\), where \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\), and the relationship \(\sin(\theta) = 0.01x\). We plotted the adjusted intensity formula for each case, which allowed us to visualize the intensity patterns on the screen as a function of x in the range of \(0\,\text{cm}\) to \(10\,\text{cm}\). The three different cases resulted in different central maximum intensities and sidelobe patterns.

Step by step solution

01

(a) \(\lambda = 10a\)

As \(\lambda = 10a\), we can substitute this into the formula for \(\beta\): \(\beta = \frac{\pi a \sin(\theta)}{10a} = \frac{\pi \sin(\theta)}{10}\) Now, let's evaluate the intensity function in this case and sketch the resulting pattern on the screen.
02

Plot the intensity as a function of \(x\)

To plot the intensity as a function of \(x\), let's first find the relationship between \(\theta\) and \(x\). Since the screen is \(1.0\,\text{m}\) away from the slit, we have \(\tan(\theta) \approx \sin(\theta) \approx \frac{x}{1.0\,\text{m}} \implies \sin(\theta) = 0.01x\) Substitute and adjust the intensity formula with this relationship: \(I(\theta) = I_0 \left[ \frac{\sin(\frac{\pi}{10}\cdot0.01x)}{\frac{\pi}{10}\cdot0.01x} \right]^2\) Now, plot this function over the range \(x=0\,\text{cm}\) to \(10\,\text{cm}\). The intensity will be maximum at the center and will have small sidelobes on either side.
03

(b) \(10\lambda=a\)

Similarly, in this case we can substitute \(10\lambda = a\) into the formula for \(\beta\): \(\beta = \frac{\pi a \sin(\theta)}{10a \lambda} = \frac{\pi a \sin(\theta)}{10a^2} = \frac{\pi \sin(\theta)}{10}\)
04

Plot the intensity as a function of \(x\)

As before, we can substitute the relationship between \(\theta\) and \(x\) in the intensity formula: \(I(\theta) = I_0 \left[ \frac{\sin(\frac{\pi}{10}\cdot0.01x)}{\frac{\pi}{10}\cdot0.01x} \right]^2\) Plot this function over the range \(x=0\,\text{cm}\) to \(10\,\text{cm}\). The intensity pattern will be different than the previous case, with the central maximum having a lower intensity and sidelobes with relatively high intensities.
05

(c) \(30\lambda=a\)

In this case, we substitute \(30\lambda=a\) into the formula for \(\beta\): \(\beta = \frac{\pi a \sin(\theta)}{30a \lambda} = \frac{\pi \sin(\theta)}{30}\)
06

Plot the intensity as a function of \(x\)

By substituting the relationship between \(\theta\) and \(x\) in the intensity formula, we have: \(I(\theta) = I_0 \left[ \frac{\sin(\frac{\pi}{30}\cdot0.01x)}{\frac{\pi}{30}\cdot0.01x} \right]^2\) Plot this function over the range \(x=0\,\text{cm}\) to \(10\,\text{cm}\). The intensity pattern will be different than the previous two cases. The central maximum will be higher in intensity, and there will be many small sidelobes on either side. In summary, plotting the adjusted intensity formula for each case allows us to visualize the intensity patterns on the screen as a function of x in the range of \(0\,\text{cm}\) to \(10\,\text{cm}\).

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Most popular questions from this chapter

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