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Chapter 16: Problem 2295

The distance between the first and sixth minima in the diffraction pattern of a single slit, it is \(0.5 \mathrm{~mm}\). The screen is \(0.5 \mathrm{~m}\) away from the Slit. If the wavelength of light is \(5000 \AA\), then the width of the slit will be \(\mathrm{mm}\) (D) \(1.0\) (A) 5 (B) \(2.5\) (C) \(1.25\)

Short Answer

Expert verified
The width of the slit is (C) \(1.25 \ \mathrm{mm}\).

Step by step solution

01

Understand the formula for minima in a single-slit diffraction pattern.

The formula for minima in a single-slit diffraction pattern is given by: \[m\lambda = a\sin\theta\]. Here, \(m\) is the order of the minima, \(\lambda\) is the wavelength of light, \(a\) is the width of the slit, and \(\theta\) is the angle formed between the central maximum and the minima. We need to find \(a\).
02

Convert the given measurements to meters.

We have been given the distance between the slit and the screen \(L = 0.5\ \text{m}\), the wavelength of light \(\lambda = 5000\ \text{\AA}\), and the distance between the first and sixth minima \(y = 0.5\ \text{mm}\). Let's convert them to meters: \(\lambda = 5000\ \text{\AA} = 5 \times 10^{-7}\ \text{m}\) \(y = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}\)
03

Calculate the angle θ using the distance between the first and sixth minima.

First, we need to find the distance between consecutive minima: \(\Delta y = \frac{5\times10^{-4}\ \text{m}}{5} = 1\times 10^{-4}\ \text{m}\). Next, since the angle \(\theta\) is small, we can use the approximation \(\tan\theta \approx \sin\theta \approx \frac{\Delta y}{L} \Rightarrow \sin\theta = \frac{1\times 10^{-4}\ \text{m}}{0.5\ \text{m}} = 2 \times 10^{-4}\).
04

Use the formula to find the width of the slit and select the correct answer.

Now, we can use the formula for minima with the given values: \(m\lambda = a\sin\theta \Rightarrow a = \frac{m\lambda}{\sin\theta}\) We are given that the distance between the first and sixth minima is 0.5 mm, which means we need to consider the distance for 5 orders (from first to the sixth minimum is a difference of 5 orders). Thus, \(m = 5\). Now, substitute the values we found: \(a = \frac{5(5 \times 10^{-7}\ \text{m})}{2 \times 10^{-4}} = \frac{25 \times 10^{-7}\ \text{m}}{2 \times 10^{-4}} = 1.25\times10^{-3}\ \text{m}\) Converting the slit width back to millimeters, we get, \(a = 1.25\ \text{mm}\). Therefore, the correct answer is (C) \(1.25\).

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Most popular questions from this chapter

A thin prism of \(3^{\circ}\), angle made from glass of refractive index \(1.5\) is combined with another thin prism made from glass of refractive index \(1.3\) to produce dispersion without deviation. What is the angle of second prism. (A) \(3^{\circ}\) (B) \(-3^{\circ}\) (C) \(-5^{\circ}\) (D) \(5^{\circ}\)

The head light of a jeep are \(1.2 \mathrm{~m}\) apart. If the pupil of the eye of an observer has a diameter of \(2 \mathrm{~mm}\) and light of wavelength \(5896 \AA\) is used what should be the maximum distance of the jeep from the observer if two head lights are just seem to be separated apart? (A) \(30.9 \mathrm{~km}\) (B) \(33.4 \mathrm{~km}\) (C) \(3.34 \mathrm{~km}\) (D) \(30.9 \mathrm{~km}\)

Light from two coherent Sources of the same amplitude \(\mathrm{A}\) and wavelength \(\lambda\), illuminates the Screen. The intensity of the central maximum is Io. If the sources were incoherent, the intensity at the same point will be (A) \(\left(\mathrm{I}_{0} / 2\right)\) (B) \(\left(\mathrm{I}_{0} / 4\right)\) (C) \(4 \mathrm{I}_{0}\) (D) \(2 \mathrm{I}_{0}\)

Ordinary light incident on a glass slab at the polarizing angle, suffers a deviation of \(22^{\circ}\). The value of angle of refraction in this case is (A) \(44^{\circ}\) (B) \(34^{\circ}\) (C) \(22^{\circ}\) (D) \(11^{\circ}\)

The two coherent sources of intensity ratio \(\beta\) produce interference. The fringe visibility will be (A) \(2 \beta\) (B) \((\beta / 2)\) (C) \(\\{\sqrt{\beta} /(1+\beta)\\}\) (D) \(\\{(2 \sqrt{\beta}) /(1+\beta)\\}\)
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