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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

A \(5.000-\mathrm{cm}\) -wide diffraction grating with 200 grooves is used to resolve two closely spaced lines (a doublet) in a spectrum. The doublet consists of two wavelengths, \(\lambda_{\mathrm{a}}=\) \(629.8 \mathrm{nm}\) and \(\lambda_{\mathrm{b}}=630.2 \mathrm{nm} .\) The light illuminates the entire grating at normal incidence. Calculate to four significant digits the angles \(\theta_{1 \mathrm{a}}\) and \(\theta_{1 \mathrm{~b}}\) with respect to the normal at which the first-order diffracted beams for the two wavelengths, \(\lambda_{\mathrm{a}}\) and \(\lambda_{\mathrm{b}}\), respectively, will be reflected from the grating. Note that this is not \(0^{\circ}\) What order of diffraction is required to resolve these two lines using this grating?

A two-slit apparatus is covered with a red \((670 \mathrm{nm})\) filter. When white light is shone on the filter, on the screen beyond the two-slit apparatus, there are nine interference maxima within the 4.50 -cm-wide central diffraction maximum. When a blue \((450 \mathrm{nm})\) filter replaces the red, how many interference maxima will there be in the central diffraction maximum, and how wide will that diffraction maximum be?

Coherent, monochromatic light of wavelength \(450.0 \mathrm{nm}\) is emitted from two locations and detected at another location. The path difference between the two routes taken by the light is \(20.25 \mathrm{~cm}\). Will the two light waves interfere destructively or constructively at the detection point?

The Michelson interferometer is used in a class of commercially available optical instruments called wavelength meters. In a wavelength meter, the interferometer is illuminated simultaneously with the parallel beam of a reference laser of known wavelength and that of an unknown laser. The movable mirror of the interferometer is then displaced by a distance \(\Delta d,\) and the number of fringes produced by each laser and passing by a reference point (a photo detector) is counted. In a given wavelength meter, a red He-Ne laser \(\left(\lambda_{\mathrm{Red}}=632.8 \mathrm{nm}\right)\) is used as a reference laser. When the movable mirror of the interferometer is displaced by a distance \(\Delta d\), a number \(\Delta N_{\text {Red }}=6.000 \cdot 10^{4}\) red fringes and \(\Delta N_{\text {unknown }}=7.780 \cdot 10^{4}\) fringes pass by the reference photodiode. a) Calculate the wavelength of the unknown laser. b) Calculate the displacement, \(\Delta d\), of the movable mirror.

Plane light waves are incident on a single slit of width \(2.00 \mathrm{~cm} .\) The second dark fringe is observed at \(43.0^{\circ}\) from the central axis. What is the wavelength of the light?
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