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Chapter 18: Problem 117

Total numbers of fringes can be seen on the screen will be (A) 2001 (B) 4001 (C) 6001 (D) 8001

Short Answer

Expert verified
The exercise does not provide enough information to determine the exact total number of fringes on the screen. To calculate the total number of fringes, we need additional information such as the distance between slits or the wavelength of the light used in the interference experiment.

Step by step solution

01

Understand the concept of fringes

Fringes are the bright and dark areas in an interference pattern. It is crucial to know how they are formed to find the total number of fringes that can be seen on the screen. In interference experiments, two or more light waves combine to form the resultant wave with varying amplitudes. The bright fringes occur where the amplitude increases due to constructive interference, while dark fringes occur due to destructive interference.
02

Determine the conditions for constructive and destructive interference

The path difference determines the occurrence of constructive and destructive interference. Constructive interference occurs when the path difference is an integral multiple of the wavelength, i.e., \(n \lambda\), where \(n\) is an integer and \(\lambda\) is the wavelength of the light. Destructive interference occurs when the path difference is an odd multiple of half the wavelength, i.e., \((2n + 1) \frac{\lambda}{2}\), where \(n\) is an integer. For a double-slit experiment, the path difference is determined by the distance between the slits (d) and the angle between the central maximum and the fringe (\(\theta\)): Path difference = d sin\(\theta\)
03

Find the number of bright fringes

To find the number of bright fringes, we need to consider the constructive interference condition, i.e., d sin\(\theta\) = \(n \lambda\). For the maximum observable angle (\(\theta = 90^{\circ}\)), the maximum number of constructive interferences will occur when sin(\(\theta\)) = 1. Therefore, d = \(n \lambda\), and the maximum value of \(n\) is the total number of bright fringes.
04

Find the number of dark fringes

Similarly, to find the number of dark fringes, we will consider the destructive interference condition, i.e., d sin\(\theta\) = \((2n + 1) \frac{\lambda}{2}\). For the maximum observable angle (\(\theta = 90^{\circ}\)), the maximum value of \(n\) will give the total number of dark fringes.
05

Find the total number of fringes

The total number of fringes will be the sum of bright and dark fringes for both sides of the central bright fringe. However, there isn't enough information in the exercise to determine the total number of fringes. The given options (A, B, C, D) indicate that the number of fringes could be one of the specified options, but without more information, we cannot calculate the distances or wavelengths necessary to narrow down the specific answer.

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Most popular questions from this chapter

A concave lens of focal length \(10 \mathrm{~cm}\) and a convex lens of focal length \(20 \mathrm{~cm}\) are placed certain distance apart. If parallel rays incident on one lens become converging after passing through other lens, then the separation between the lenses must be greater than (A) Zero (B) \(5 \mathrm{~cm}\) (C) \(10 \mathrm{~cm}\) (D) \(9 \mathrm{~cm}\)

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) \(1.75 \mathrm{~mm}\)

A microscope is focused on a needle lying in an empty tank. Now, the tank is filled with benzene to a height \(120 \mathrm{~mm}\). The microscope is moved \(40 \mathrm{~mm}\) to focus the needle again. The refractive index of benzene is (A) \(1.5\) (B) \(2.5\) (C) \(3.0\) (D) \(4.5\)

Interference fringes were produced in Young's double slit experiment using light of wavelength \(5000 \AA\). When a film of thickness \(2.5 \times 10^{-3} \mathrm{~cm}\) was placed in front of one of the slits, the fringe pattern shifted by a distance equal to 20 fringe-widths. The refractive index of the material of the film is (A) \(1.25\) (B) \(1.35\) (C) \(1.4\) (D) \(1.5\)

If \(\varepsilon_{0}\) and \(\mu_{0}\) represent the permittivity and permeability of vacuum, \(\varepsilon\) and \(\mu\) represent the permittivity and permeability of medium, then refractive index of the medium is given by (A) \(\sqrt{\frac{\mu_{0} \varepsilon_{0}}{\mu \varepsilon}}\) (B) \(\sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}}\) (C) \(\sqrt{\frac{\varepsilon}{\mu_{0} \varepsilon_{0}}}\) (D) \(\sqrt{\frac{\mu_{0} \varepsilon_{0}}{\mu}}\)
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