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Chapter 18: Problem 61

In Young's double slit experiment, when sodium light of wavelength \(5893 \AA\) is used, then 62 fringes are seen in the field of view. Instead of sodium light, if violet light of wavelength \(4358 \AA\) is used, then the number of fringes that will be seen in the field of view will be (A) 54 (B) 64 (C) 74 (D) 84

Short Answer

Expert verified
The number of fringes for violet light (n_violet) is calculated using the relationship \(n_{\text{sodium}} / n_{\text{violet}} = \lambda_{\text{sodium}} / \lambda_{\text{violet}}\). After converting wavelengths to meters and plugging in the given values, we find that \(n_{\text{violet}} \approx 45.71\). Since the number of fringes must be a whole number, we round up to 46 fringes, which is not among the available options (A, B, C, D).

Step by step solution

01

Find the relationship between the number of fringes and wavelength

The number of fringes in the field of view can be determined using the formula: n = 2d * sin(胃) / 位 Where n is the number of fringes, d is the distance between the slits, 胃 is the angle of deviation, and 位 is the wavelength of the light. Notice that this formula depends on the wavelength, so if we find the ratio of fringes (n) for the sodium light and violet light, we can find a relationship between them: n_sodium / n_violet = 位_sodium / 位_violet
02

Convert given wavelengths into meters

It's essential to convert the wavelengths given in Angstroms into meters, as we'll work in SI units: 位_sodium = 5893 * 10^-10 m 位_violet = 4358 * 10^-10 m
03

Find the ratio of fringes for sodium and violet light

Now we can find the ratio of fringes for sodium and violet light using the relationship we derived in step 1: n_sodium / n_violet = (5893 * 10^-10) / (4358 * 10^-10) n_sodium / n_violet = 5893 / 4358
04

Calculate the number of fringes for violet light

We are given the number of fringes for sodium light (n_sodium = 62). Now, we'll solve for the number of fringes for the violet light (n_violet): n_violet = n_sodium * (4358 / 5893) n_violet = 62 * (4358 / 5893) n_violet 鈮� 45.71 Since the number of fringes must be a whole number, we will round up to 46 fringes. This means the answer is not in the available options (A, B, C, D). Nevertheless, this is the method to appropriately solve the problem.

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Most popular questions from this chapter

A plano-convex lens has a thickness of \(4 \mathrm{~cm}\). When placed on a horizontal table with curved surface in contact with it, the apparent depth of the bottom-most point of the lens is found to be \(3 \mathrm{~cm}\). If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of plane face is found to be \(25 / 8 \mathrm{~cm}\). The focal length of the lens is (A) \(50 \mathrm{~cm}\) (B) \(75 \mathrm{~cm}\) (C) \(100 \mathrm{~cm}\) (D) \(150 \mathrm{~cm}\)

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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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