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Chapter 25: Problem 65

When a double slit is illuminated with light of wavelength \(510 \mathrm{nm},\) the interference maxima on a screen \(2.4 \mathrm{m}\) away gradually decrease in intensity on either side of the 2.40 -cm-wide central maximum and reach a minimum in a spot where the fifth-order maximum is expected. (a) What is the width of the slits? (b) How far apart are the slits?

Short Answer

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Question: In a double-slit interference experiment, the wavelength of light used is 510 nm. The distance between the central maximum and the point where the 5th-order maximum is expected is 2.4 cm, and the distance between the screen and the slits is 2.4 m. Determine (a) the width of the slits and (b) the distance between the slits. Answer: (a) The width of the slits is approximately 51 μm. (b) The distance between the slits is approximately 510 μm.

Step by step solution

01

Write down the given information

We are given the following information: - Wavelength of light, \(\lambda=510\) nm or \(5.1\times10^{-7}\) m. - Distance between the central maximum and the minimum where the 5th-order maximum is expected, \(Y_{5th}=2.4 \times10^{-2}\) m. - Distance between the slits and the screen, \(L=2.4\) m.
02

Calculate the angle for the 5th-order maximum, \(\theta_{5th}\)

We can use the formula relating the distance where the 5th order maximum is expected to the screen and slit distance with the angle: $$ \tan(\theta_{5th}) = \frac{Y_{5th}}{L} $$ Now we can calculate the angle, \(\theta_{5th}\): $$ \theta_{5th} = \tan^{-1}\left(\frac{Y_{5th}}{L}\right) = \tan^{-1}\left(\frac{2.4 \times10^{-2} \mathrm{m}}{2.4 \mathrm{m}}\right) $$ So, \(\theta_{5th} \approx 0.01\) radians.
03

Use the minima condition to find the slit width, \(w\)

For the minimum intensity in the interference pattern (dark spot), the following condition must be met: $$ w\sin(\theta) = m\lambda, \quad m=1,2,3,\ldots $$ We have minimum intensity at the 5th-order maximum, so we can set \(m=5\) and \(\theta = \theta_{5th}\) to find the slit width, \(w\): $$ w = \frac{5\lambda}{\sin(\theta_{5th})} = \frac{5(5.1\times10^{-7}\,\mathrm{m})}{\sin(0.01)} $$ Therefore, \(w \approx 5.1 \times 10^{-5}\) m or \(51\,\mathrm{\mu m}\).
04

Calculate the distance between adjacent maxima, \(\Delta y\)

The formula relating the distance between adjacent maxima (\(\Delta y\)), the wavelength (\(\lambda\)), the distance between slits and the screen (\(L\)), and the distance between the slits (\(d\)) is: $$ \Delta y = \frac{\lambda L}{d} $$ We know the distance between the first maximum and the central maximum (which are also adjacent maxima) is the same as \(Y_{5th}\). So we can set \(\Delta y = Y_{5th} = 2.4 \times 10^{-2} \,\mathrm{m}\) and solve for \(d\): $$ d = \frac{\lambda L}{\Delta y} = \frac{(5.1\times10^{-7}\,\mathrm{m})(2.4\,\mathrm{m})}{2.4 \times10^{-2}\,\mathrm{m}} $$ Therefore, \(d \approx 5.1 \times 10^{-4}\) m or \(510\,\mathrm{\mu m}\).
05

Answers

(a) The width of the slits is approximately \(51\,\mathrm{\mu m}\). (b) The distance between the slits is approximately \(510\,\mathrm{\mu m}\).

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