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Chapter 18: Problem 12

A thin sheet of glass \((\mu=1.5)\) of thickness 6 microns introduced in the path of one of interfering beams of a double slit experiment shifts the central fringes to a position previously occupied by fifth bright fringe. Then the wavelength of the light used is (A) \(6000 \AA\) (B) \(3000 \AA\) (C) \(4500 \AA\) (D) \(7500 \AA\)

Short Answer

Expert verified
The wavelength of the light used in the double-slit experiment is \(6000 \AA\). The correct answer is (A) \(6000 \AA\).

Step by step solution

01

Optical Path Difference

Recall that the optical path difference is given by the formula: Optical Path Difference = (碌 - 1) * thickness. In this problem, we have 碌 = 1.5 and the thickness of the glass = 6 microns. Let's calculate the optical path difference: Optical Path Difference = (1.5 - 1) * 6 microns = 0.5 * 6 microns = 3 microns
02

Relationship between Optical Path Difference and Fringe Shift

We are told that the central fringe shifts to the position previously occupied by the fifth bright fringe. This means that the optical path difference caused by the glass must be equal to five times the wavelength (5位) of the light used in the double-slit experiment: Optical Path Difference = 5位
03

Solve for Wavelength

Now we have enough information to solve for the wavelength of the light: 3 microns = 5位 To find the wavelength, divide both sides of the equation by 5: 位 = (3 microns) / 5 = 0.6 microns
04

Convert Wavelength to Angstroms

Finally, we need to convert the wavelength from microns to Angstroms. Recall that 1 micron = 10,000 Angstroms: 位 = 0.6 microns * 10,000 Angstroms/micron = 6000 Angstroms The wavelength of the light used in the double-slit experiment is 6000 Angstroms. The correct answer is (A) \(6000 \AA\).

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Most popular questions from this chapter

A thin film of thickness \(t\) and index of refraction \(1.33\) coats a glass with index of refraction 1.50. Which of the following thickness \(t\) will not reflect normally incident light with wavelength \(640 \mathrm{~nm}\) in air? (A) \(120 \mathrm{~nm}\) (B) \(240 \mathrm{~nm}\) (C) \(300 \mathrm{~nm}\) (D) \(480 \mathrm{~nm}\)

For which of the pairs of \(u\) and \(f\) for a mirror image is smaller in size than the object (A) \(u=-10 \mathrm{~cm}, f=20 \mathrm{~cm}\) (B) \(u=-20 \mathrm{~cm}, f=-30 \mathrm{~cm}\) (C) \(u=-45 \mathrm{~cm}, f=-10 \mathrm{~cm}\) (D) \(u=-60 \mathrm{~cm}, f=30 \mathrm{~cm}\)

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

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 an interference arrangement, similar to Young's double slit experiment, the slits \(S_{1}\) and \(S_{2}\) are illuminated with coherent microwave sources, each of frequency \(10^{6} \mathrm{~Hz}\). The sources are synchronized to have zero phase difference. The slits are separated by distance \(d=150.0 \mathrm{~m}\). The intensity \(I_{(\theta)}\) is measured as a function of \(\theta\), where \(\theta\) is defined as shown in the Fig. 18.53. If \(I_{0}\) is maximum intensity, then \(I_{(\theta)}\) for \(0 \leq \theta \leq 90\) is given by (A) \(I_{(\theta)}=I_{0}\) for \(\theta=0^{\circ}\) (B) \(I_{(\theta)}=\left(I_{0} / 2\right)\) for \(\theta=30^{\circ}\) (C) \(I_{(\theta)}=\left(I_{0} / 4\right)\) for \(\theta=90^{\circ}\) (D) \(I_{(\theta)}\) is constant for all values of \(\theta\)
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