Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Monochromatic light (wavelength $= 450 n m$ ) is incident perpendicularly on a single slit (width $= 0.4 m m$ ). A screen is placed parallel to the slit plane, and on it the distance between the two minima on either side of the central maximum is $1.8 m m$ .
**(a) What is the distance from the slit to the screen? (Hint: The angle to either minimum is small enough that $s i n θ \approx t a n θ$ .)
(b) What is the distance on the screen between the first minimum and the third minimum on the same side of the central maximum?**

(a) The distance of the screen from the slit is $0.8 m$ .

(b) The distance between the first and third minima is $1.8 m m$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Slit width $a = 0.4 m m$

Distance between the two minima on either side of the central maxima localid="1663048702619" $∆ y_{1, - 1} = 1.8 m m$

Wavelength of incident light $λ = 450 n m$

Step 2: Definition and concept of diffraction from a grating

An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.

The angular distance $θ$ of the $m_{t h}$ order minima in diffraction pattern produced from a single slit having slit width $a$ is

$a s i n θ = m λ$ …(i)

Here, $λ$ is the wavelength of the incident light.

Step 3: (a) Determining the distance of the screen from the slit

Let the distance from the slit to the screen be $D$ . For small angular distances

$\sin θ \approx \tan θ = \frac{y}{D}$

Here $y$ is the distance measured on the screen. Thus, from equation (i) the separation between the first two minima $m \pm 1$ on either sides of the central maxima is

$\frac{a ∆ y_{1, - 1}}{D} = \{1 - (- 1)\} λ D = \frac{a ∆ y_{1, - 1}}{2 λ}$

Substitute the values to get

$D = \frac{0.4 m m \times 1.8 m m}{2 \times 450 n m} = \frac{0.4 \times 10^{- 3} m \times 1.8 \times 10^{- 3} m}{2 \times 450 \times 10^{- 9} m} = 0.8 m$

Thus, the distance is $0.8 m$ .

Step 4: (b) Determining the distance between the first and third minima

From equation (i), the distance between the first $(m = 1)$ and third minima $(m = 3)$ is

$\frac{a ∆ y_{1, 3}}{D} = \{3 - 1\} λ ∆ y_{1, 3} = \frac{2 λ D}{a} = \frac{2 \times 450 \times 10^{- 9} m \times 0.8 m}{0.4 \times 10^{- 3} m} = 1.8 m m$

Thus, the distance is $1.8 m m$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Definition and concept of diffraction from a grating

Step 3: (a) Determining the distance of the screen from the slit

Step 4: (b) Determining the distance between the first and third minima

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