/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 One slit in a YDSE set up is cov... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 93

One slit in a YDSE set up is covered by a glass plate (Refractive index \(=\mu_{1}\) ) and other by another glass plate (Refractive index \(=\mu_{2}\) ) of same thickness. If \(I_{0}\) is the intensity of light through each slit and \(\lambda\) is the wavelength of light, then intensity at point \(P\) is ( \(P\) is symmetrical with respect to slits \(S_{1}\) and \(S_{2}\) ) (A) \(4 I_{0} \cos ^{2} \frac{\pi}{\lambda}\left(\mu_{1}-\mu_{2}\right) t\) (B) \(4 I_{0} \cos ^{2} \frac{\pi}{\lambda}\left(\mu_{1}+\mu_{2}\right) t\) (C) \(4 I_{0} \sin \frac{\pi}{\lambda}\left(\mu_{1}-\mu_{2}\right) t\) (D) \(4 I_{0} \sin \frac{\pi}{\lambda}\left(\mu_{1}+\mu_{2}\right) t\)

Short Answer

Expert verified
The correct answer is (A) \(4 I_{0} \cos ^{2} \frac{\pi}{\lambda}\left(\mu_{1}-\mu_{2}\right) t\).

Step by step solution

01

- Calculation of the optical path difference

As the first step, calculate the optical path difference due to the two glass plates. The optical path length for a wave traveling through a medium is given by the product of the physical length and the refractive index. Therefore, for glass plates with identical physical thickness t but differing refractive indices \(\mu_1\) and \(\mu_2\), the optical path difference will be \(\Delta = t(\mu_{1}-\mu_{2})\).
02

- Calculating the phase difference

The phase difference (\( \delta \)) between the waves from the two slits is directly proportional to the path difference. One wavelength of light corresponds to a phase difference of \(2\pi\), and therefore \(\delta = 2\pi \cdot \frac{\Delta}{\lambda}\), which is equal to \(2\pi \cdot \frac{t(\mu_{1}-\mu_{2})}{\lambda}\).
03

- Calculating the result intensity at point P

We know that the intensity at a point is the square of the amplitude of the resultant wave. In case of interference, the resultant intensity at a point P is given by \(I = 4I_0 \cos^2(\frac{\delta}{2})\). Substitute the value of \(\delta\) that we got from step 2 into this equation. After this substitution, the intensity \(I\) at point \(P\) simplifies to \(I = 4I_0 \cos^2(\frac{\pi}{\lambda} (\mu_{1}-\mu_{2})t)\).

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Most popular questions from this chapter

A circular beam of light of diameter \(d=2 \mathrm{~cm}\) falls on a plane surface of glass. The angle of incidence is \(60^{\circ}\) and refractive index of glass is \(\mu=3 / 2\). The diameter of the refracted beam is (A) \(4.0 \mathrm{~cm}\) (B) \(3.0 \mathrm{~cm}\) (C) \(3.26 \mathrm{~cm}\) (D) \(2.52 \mathrm{~cm}\)

When an unpolarized light of intensity \(I_{0}\) is incident on a polarizing sheet, the intensity of the light which does not get transmitted is (A) \(\frac{1}{4} I_{0}\) (B) \(\frac{1}{2} I_{0}\) (C) \(I_{0}\) (D) Zero

A plastic hemisphere has a radius of curvature of \(8 \mathrm{~cm}\) and an index of refraction of \(1.6 .\) On the axis half way between the plane surface and the spherical one ( \(4 \mathrm{~cm}\) from each) is a small object \(O .\) The distance between the two images when viewed along the axis from the two sides of the hemisphere is approximately. (A) \(1.0 \mathrm{~cm}\) (B) \(1.5 \mathrm{~cm}\) (C) \(3.75 \mathrm{~cm}\) (D) \(2.5 \mathrm{~cm}\)

Two polaroids are kept crossed to each other. Now one of them is rotated through an angle of \(45^{\circ}\). The percentage of unpolarized incident light now transmitted through the system is (A) \(15 \%\) (B) \(25 \%\) (C) \(50 \%\) (D) \(60 \%\)

Two beams, \(A\) and \(B\), of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam \(A\) has maximum intensity (and beam \(B\) has zero intensity), a rotation of polaroid through \(30^{\circ}\) makes the two beams appear equally bright. If the initial intensities of the two beams are \(I_{A}\) and \(I_{B}\), respectively, then \(\frac{I_{A}}{I_{B}}\) equals [2014] (A) 3 (B) \(3 / 2\) (C) 1 (D) \(1 / 3\)
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