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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

The ratio of intensities of rays emitted from two different coherent Sourees is \(\alpha .\) For the interference pattern by them, $\left[\left(I_{\max }+I_{\min }\right) /\left(I_{\max }-I_{\min }\right)\right]$ will be equal to (A) \(\\{(1+\sqrt{\alpha}) / 2 \alpha\\}\) (B) \(\\{(1+\alpha) / 2 \alpha\\}\) (C) \(\\{(1+\sqrt{\alpha}) / 2\\}\) (D) \(\\{(1+\alpha) /(2 \sqrt{\alpha})\\}\)

A convex lens of glass \((\mathrm{n}=1.5)\) has focal length \(0.2 \mathrm{~m}\). The lens is immersed in water of refractive index \(1.33\). The change in the power of convex lens is (A) \(3.72 \mathrm{D}\) (B) \(4.62 \mathrm{D}\) (C) \(6.44 \mathrm{D}\) (D) \(1.86 \mathrm{D}\)

For four different transparent medium $\mathrm{n}_{41} \times \mathrm{n}_{12} \times \mathrm{n}_{21}=$ (A) \(\left(1 / \mathrm{n}_{41}\right)\) (B) \(\mathrm{n}_{41}\) (C) \(\mathrm{n}_{14}\) (D) \(\left(1 / \mathrm{n}_{14}\right)\)

What is the time taken in seconds to cross a glass plate of thickness $6 \mathrm{~mm}\( and \)\mu=2.0$ by light ? (A) \(8 \times 10^{-11}\) (B) \(4 \times 10^{-11}\) (C) \(2 \times 10^{11}\) (D) \(16 \times 10^{-11}\)

A sound source emits sound of \(600 \mathrm{~Hz}\) frequency, this sound enters by opened door of width \(0.75 \mathrm{~m} .\) Find the angle on one side at which first minimum is formed. The speed of sound \(=300 \mathrm{~ms}^{-1}\) (A) \(84.4^{\circ}\) (B) \(90^{\circ}\) (C) \(74.2^{\circ}\) (D) \(41.2^{\circ}\)
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