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Chapter 34: Problem 25

Coherent monochromatic light with wavelength \(\lambda=514 \mathrm{nm}\) is incident on two slits that are separated by a distance \(d=0.500 \mathrm{~mm} .\) The intensity of the radiation at a screen \(2.50 \mathrm{~m}\) away from each slit is \(180.0 \mathrm{~W} / \mathrm{cm}^{2} .\) Deter-

Short Answer

Expert verified
Answer: The intensity of light at an angle of \(30^{\circ}\) from the central maximum is \(0 \, \mathrm{W/cm}^2\).

Step by step solution

01

Calculate the path difference at \(30^{\circ}\)

To find the path difference, we can use the formula \(d \, \sin{\theta}\), where \(d\) is the distance between the slits and \(\theta\) is the given angle. Here, \(d=0.500 \mathrm{~mm}\) and \(\theta = 30^{\circ}\). Calculate the path difference: $$ \Delta{x} = d \, \sin{\theta} = 0.500 \times 10^{-3}\, \mathrm{m} \times \sin{30^{\circ}} = 0.250 \times 10^{-3}\, \mathrm{m}. $$
02

Calculate the phase difference

To calculate the phase difference, we can use the formula \(\delta = \dfrac{2\pi}{\lambda} \Delta{x}\), where \(\lambda = 514 \,\mathrm{nm}\). Calculate the phase difference: $$ \delta = \dfrac{2\pi}{514 \times 10^{-9}\, \mathrm{m}} \times 0.250 \times 10^{-3}\, \mathrm{m} = \dfrac{2\pi}{514 \times 10^{-9}\, \mathrm{m}} \times 250 \times 10^{-9}\, \mathrm{m} = \pi. $$
03

Calculate the resultant amplitude

To obtain the resultant amplitude, we can apply the phase difference to the amplitude formula: $$ A_r = 2A\cos{\dfrac{\delta}{2}} = 2A\cos{\dfrac{\pi}{2}} = 2A \times 0 = 0, $$ where \(A_r\) is the resultant amplitude and \(A\) is the amplitude of each wave from the two slits. The resultant amplitude turns out to be zero.
04

Calculate the intensity at \(30^{\circ}\) angle

Finally, we can find the intensity using the formula \(I = kA_r^2\), where \(k\) is a proportionality constant. Since the resultant amplitude is zero, the intensity will also be zero: $$ I = k \times 0^2 = 0. $$ At an angle of \(30^{\circ}\) from the central maximum, the intensity is \(0 \, \mathrm{W/cm}^2\).

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Most popular questions from this chapter

What minimum path difference is needed to cause a phase shift by \(\pi / 4\) in light of wavelength \(700 . \mathrm{nm} \)

A laser beam with wavelength \(633 \mathrm{nm}\) is split into two beams by a beam splitter. One beam goes to Mirror \(1,\) a distance \(L\) from the beam splitter, and returns to the beam splitter, while the other beam goes to Mirror \(2,\) a distance \(L+\Delta x\) from the beam splitter, and returns to the same beam splitter. The beams then recombine and go to a detector together. If \(L=1.00000 \mathrm{~m}\) and \(\Delta x=1.00 \mathrm{~mm},\) which best describes the kind of interference at the detector? (Hint: To doublecheck your answer, you may need to use a formula that was originally intended for combining two beams in a different geometry, but which still is applicable here.) a) purely constructive b) purely destructive c) mostly constructive d) mostly destructive e) neither constructive nor destructive

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