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Chapter 25: Problem 49

Light from a red laser passes through a single slit to form a diffraction pattern on a distant screen. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum on the screen?

Short Answer

Expert verified
Answer: The width of the central maximum becomes half of its original value when the width of the slit is increased by a factor of two.

Step by step solution

01

Understand the basics of single-slit diffraction patterns

Single-slit diffraction is an interference pattern formed when light passes through a single, narrow slit and spreads out in a series of bright and dark fringes on a distant screen. The central maximum is the brightest part of the pattern, and its width depends on the width of the slit and the wavelength of the light.
02

Use the formula for the angular width of the central maximum

The angular width (胃) of the central maximum in a single-slit diffraction pattern is given by the formula: 胃 鈮� \(\frac{2位}{a}\) where 位 is the wavelength of the light and a is the width of the slit.
03

Calculate the new angular width of the central maximum with the increased slit width

Let's denote the original slit width as a and the new slit width(after increasing by a factor of two) as a'. Then, a' = 2a Now, we will plug this new slit width into the formula for the angular width of the central maximum: 胃' 鈮� \(\frac{2位}{a'}\) 胃' 鈮� \(\frac{2位}{2a}\) 胃' 鈮� \(\frac{位}{a}\)
04

Compare the new angular width with the original angular width

Now, let's compare the new angular width (胃') with the original angular width (胃). We have, 胃 鈮� \(\frac{2位}{a}\) and 胃' 鈮� \(\frac{位}{a}\) Dividing 胃' by 胃, we get: \(\frac{胃'}{胃}\) = \(\frac{\frac{位}{a}}{\frac{2位}{a}}\) \(\frac{胃'}{胃}\) = \(\frac{1}{2}\) This means that the width of the central maximum on the screen will be half of what it originally was when the width of the slit is increased by a factor of two.

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Most popular questions from this chapter

A pinhole camera doesn't have a lens; a small circular hole lets light into the camera, which then exposes the film. For the sharpest image, light from a distant point source makes as small a spot on the film as possible. What is the optimum size of the hole for a camera in which the film is $16.0 \mathrm{cm}$ from the pinhole? A hole smaller than the optimum makes a larger spot since it diffracts the light more. A larger hole also makes a larger spot because the spot cannot be smaller than the hole itself (think in terms of geometrical optics). Let the wavelength be \(560 \mathrm{nm}\).
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