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Chapter 18: Problem 148

In Young's double slit experiment, the two slits act as coherent sources of equal amplitude \(A\) and wavelength \(\lambda\). In another experiment with the same set-up, the two slits are source of equal amplitude \(A\) and wavelength \(\lambda\), but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that of second case is

Short Answer

Expert verified
The ratio of the intensity of light at the midpoint of the screen in the first case (coherent case) to that of the second case (incoherent case) is 2.

Step by step solution

01

Understand coherent and incoherent sources

Coherent sources are sources that emit light waves with a constant phase difference. When these waves interfere, they produce a clear interference pattern. Incoherent sources are sources that emit light waves without a constant phase difference. When these waves interfere, they don't produce any clear interference pattern.
02

Calculate the intensity for the coherent case

For the coherent case, let the amplitudes of the waves from the two slits be \(A\) and \(A\). The resulting amplitude at the midpoint of the screen would be \(A_{net} = A + A = 2A\). This is because at the midpoint, the waves from the two slits will have traveled the same distance, so they hence have the same phase and interfere constructively. The intensity of a wave is proportional to the square of its amplitude. So, the intensity at the midpoint of the screen for coherent case is given by, \(I_{coherent} = k\cdot|A_{net}|^2\), where k is the proportionality constant. Hence, we have: \(I_{coherent} = k\cdot(2A)^2 = 4kA^2\).
03

Calculate the intensity for the incoherent case

For the incoherent case, since the interference is not constructive, the intensities from each slit will simply add up at the midpoint. The intensity from each slit, \(I_{single} = k\cdot|A|^2 = kA^2\). Now, due to the incoherent nature of the light, we simply add the intensity from each slit separately at the center: \(I_{incoherent} = I_{single} + I_{single} = 2kA^2\).
04

Find the ratio of intensities

Now, we have the intensity of light in both coherent and incoherent cases. We need to find the ratio of these intensities: \(\frac{I_{coherent}}{I_{incoherent}}\) \(\frac{I_{coherent}}{I_{incoherent}} = \frac{4kA^2}{2kA^2} = \frac{4}{2} = 2\) So, the ratio of the intensity of light at the midpoint of the screen in the first case to that of the second case is 2.

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Most popular questions from this chapter

An astronomical telescope has magnifying power 10 . The focal length of the eyepiece is \(20 \mathrm{~cm}\). The focal length of the objective is (A) \(\frac{1}{200} \mathrm{~cm}\) (B) \(\frac{1}{2} \mathrm{~cm}\) (C) \(2 \mathrm{~cm}\) (D) \(200 \mathrm{~cm}\)

In YDSE distance between the slits plane and screen is \(1 \mathrm{~m}\) and distance between two slits is \(5 \mathrm{~mm}\). If slabs of thickness \(2 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) having refractive index \(1.5\) and \(1.4\) are placed in front of two slits, the shift of central maximum will be (A) \(2 \mathrm{~m}\) (B) \(8 \mathrm{~cm}\) (C) \(20 \mathrm{~cm}\) (D) \(80 \mathrm{~cm}\)

The box of a pin hole camera, of length \(L\), has a hole of radius \(a\). It is assumed that when the hole is illuminated by a parallel beam of light of wavelength the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say \(b_{\min }\) ) when \([2016]\) (A) \(a=\sqrt{\lambda L}\) and \(b_{\min }=\left(\frac{2 \lambda^{2}}{L}\right)\) (B) \(a=\sqrt{\lambda L}\) and \(b_{\min }=\sqrt{4 \lambda L}\) (C) \(a=\frac{\lambda^{2}}{L}\) and \(b_{\min }=\sqrt{4 \lambda L}\) (D) \(a=\frac{\lambda^{2}}{L}\) and \(b_{\min }=\left(\frac{2 \lambda^{2}}{L}\right)\)

A fish looks up through the water sees the outside world contained in a circular horizon. If the refractive index of water is \(4 / 3\) and the fish is \(12 \mathrm{~cm}\) below the surface, the radius of this circle in \(\mathrm{cm}\) is \(\quad\) [2005] (A) \(\frac{36}{\sqrt{7}}\) (B) \(36 \sqrt{7}\) (C) \(4 \sqrt{5}\) (D) \(36 \sqrt{5}\)

The magnifying power of an astronomical telescope in normal adjustment is 8 and the distance between the two lenses is \(54 \mathrm{~cm}\). The focal length of eye lens and objective lens will be, respectively. (A) \(6 \mathrm{~cm}\) and \(48 \mathrm{~cm}\) (B) \(48 \mathrm{~cm}\) and \(6 \mathrm{~cm}\) (C) \(8 \mathrm{~cm}\) and \(64 \mathrm{~cm}\) (D) \(64 \mathrm{~cm}\) and \(8 \mathrm{~cm}\)
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