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Chapter 3: Problem 65

Eight slits equally separated by \(0.149 \mathrm{mm}\) is uniformly illuminated by a monochromatic light at \(\lambda=523 \mathrm{nm}\) What is the width of the central principal maximum on a screen \(2.35 \mathrm{m}\) away?

Short Answer

Expert verified
The width of the central principal maximum on the screen is approximately \(87.88 \mathrm{mm}\).

Step by step solution

01

1. Find the slit spacing (d)

We are given the distance between the slits to be \(0.149\mathrm{mm}\). Let's convert it to meters: \[d = 0.149\times10^{-3} \mathrm{m}\]
02

2. Calculate the angular width of the central maximum

The angular width of the central maximum for a multiple slit grating is given by the formula: \[\Delta\theta = \frac{2\lambda}{N d}\] where \(\Delta\theta\) is the angular width of the central maximum, \(N\) is the number of slits, \(d\) is the slit spacing, and \(\lambda\) is the wavelength of the light. We are given \(N=8\), \(d = 0.149\times10^{-3}\mathrm{m}\), and \(\lambda = 523\times10^{-9}\mathrm{m}\). Substituting these values, we can find the angular width of the central maximum: \[\Delta\theta = \frac{2\times(523\times10^{-9}\mathrm{m})}{8\times(0.149\times10^{-3}\mathrm{m})}\]
03

3. Calculate the linear width of the central maximum on the screen

The linear width of the central maximum on the screen, denoted by \(W\), can be found by considering the distant L between the grating and the screen: \[W = L\times\tan\Delta\theta\] Given the distance to the screen is \(L = 2.35\mathrm{m}\), we can find the width of the central maximum: \[W = 2.35\mathrm{m} \times \tan\left(\frac{2\times(523\times10^{-9}\mathrm{m})}{8\times(0.149\times10^{-3}\mathrm{m})}\right)\]
04

4. Convert the width to millimeters

Finally, we can convert the linear width of the central maximum back to millimeters to obtain the final result: \[W_{mm} = W\times10^3 \] Calculating the width using the given values, we get \(W_{mm} = 87.88 \mathrm{mm}\). Therefore, the width of the central principal maximum on the screen is approximately \(87.88 \mathrm{mm}\).

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Most popular questions from this chapter

A good quality camera "lens" is actually a system of lenses, rather than a single lens, but a side effect is that a reflection from the surface of one lens can bounce around many times within the system, creating artifacts in the photograph. To counteract this problem, one of the lenses in such a system is coated with a thin layer of material \((n=1.28)\) on one side. The index of refraction of the lens glass is \(1.68 .\) What is the smallest thickness of the coating that reduces the reflection at \(640 \mathrm{nm}\) by destructive interference? (In other words, the coating's effect is to be optimized for \(\lambda=640 \mathrm{nm} .)\)

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