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Chapter 18: Problem 73

When a thin transparent sheet of refractive index \(\mu=\frac{3}{2}\) is placed near one of the slits in Young's double slits experiment, the intensity at the centre of the screen reduces to half of the maximum intensity. The minimum thickness of the sheet should be (A) \(\frac{\lambda}{4}\) (B) \(\frac{\lambda}{8}\) (C) \(\frac{\lambda}{2}\) (D) \(\frac{\lambda}{3}\)

Short Answer

Expert verified
The minimum thickness of the sheet should be \(\frac{\lambda}{3}\). So, the correct option is (D).

Step by step solution

01

Understanding Intensity and Amplitude Relationship

The intensity \(I\) at any point in an interference pattern is given by \(I=I_1+I_2+2\sqrt{I_1I_2}\cos \delta\), where \(I_1\) and \(I_2\) are the intensities of the two interfering waves and \(\delta\) is the phase difference between them. Given the intensity at the center is half the maximum intensity (\(I_0\)), this indicates a phase difference of \(\pi\) or 180 degrees. This is because the light from the thin film undergoes a phase shift as it passes through and emerges out of the film. To find the phase difference, we use the formula \(\delta = \frac{2\pi}{\lambda}\Delta x\), where \(\Delta x\) is the optical path difference.
02

Calculating Optical Path Difference

The optical path difference \(\Delta x\) is given by the product of the refractive index \(\mu\) and the thickness \(t\) of the film, i.e., \(\mu t\). Our phase difference \(\delta = \pi\), so substituting and solving for \(t\) gives \(\Delta x = \frac{\delta \lambda}{2 \pi} = \frac{\pi \lambda}{2 \pi} = \frac{\lambda}{2}\). Consequently, \(\mu t = \frac{\lambda}{2}\).
03

Calculate Thickness of Sheet

Using the provided value for the refractive index, \(\mu = \frac{3}{2}\), calculate for the thickness \(t\) of the thin film. This gives \(t = \frac{\frac{\lambda}{2}}{\mu} = \frac{\frac{\lambda}{2}}{\frac{3}{2}} = \frac{\lambda}{3}\).

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Most popular questions from this chapter

The refractive index of water is \(4 / 3\) and that of glass is \(5 / 3\). Then the critical angle for a ray of light entering in water from glass will be (A) \(\sin ^{-1}\left(\frac{4}{5}\right)\) (B) \(\sin ^{-1}\left(\frac{5}{4}\right)\) (C) \(\sin ^{-1}\left(\frac{20}{9}\right)\) (D) \(\sin ^{-1}\left(\frac{9}{20}\right)\)

Two convex lenses placed in contact form the image of a distant object at \(P\). If the lens \(B\) is moved to the right, the image will (A) move to the left. (B) move to the right. (C) remain at \(P\). (D) move either to the left or right, depending upon focal lengths of the lenses.

In a Young's double slit experiment, the slit separation is \(1 \mathrm{~mm}\) and the screen is \(1 \mathrm{~m}\) from the slit. For a monochromatic light of wavelength \(500 \mathrm{~nm}\), the distance of third minimum from the central maximum is (A) \(0.50 \mathrm{~mm}\) (B) \(1.25 \mathrm{~mm}\) (C) \(1.50 \mathrm{~mm}\) (D) \(1.75 \mathrm{~mm}\)

An object is placed at \(20 \mathrm{~cm}\) from a convex mirror of focal length \(10 \mathrm{~cm}\). The image formed by the mirror is (A) real and at \(20 \mathrm{~cm}\) from the mirror. (B) virtual and at \(20 \mathrm{~cm}\) from the mirror. (C) virtual and at \((20 / 3) \mathrm{cm}\) from the mirror. (D) real and at ( \(20 / 3) \mathrm{cm}\) from the mirror.

In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern (A) The intensities of both the maximum and minimum increase (B) The intensity of the maximum increases and the minimum has zero intensity (C) The intensity of the maximum decreases and that of minimum increases (D) The intensity of the maximum decreases and the minimum has zero intensity
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