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Expert-verified Found in: Page 1115 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Monochromatic light (wavelength${\mathbf{=}}{\mathbf{450}}{\mathbf{}}{\mathbit{n}}{\mathbit{m}}$ ) is incident perpendicularly on a single slit (width${\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{4}}{\mathbf{}}{\mathbit{m}}{\mathbit{m}}$ ). 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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{8}}{\mathbit{m}}{\mathbit{m}}$ . (a) What is the distance from the slit to the screen? (Hint: The angle to either minimum is small enough that${\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbit{\theta }}{\mathbf{\approx }}{\mathbit{t}}{\mathbit{a}}{\mathbit{n}}{\mathbit{\theta }}$ .) (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$0.8m$ .

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

See the step by step solution

## Step 1: Given data

Slit width $a=0.4mm$

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

Wavelength of incident light $\lambda =450nm$

## 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 $\theta$ of the ${m}_{th}$ order minima in diffraction pattern produced from a single slit having slit width $a$ is

${\mathbit{a}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbit{\theta }}{\mathbf{=}}{\mathbit{m}}{\mathbit{\lambda }}$ …(i)

Here, $\lambda$ 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

$\mathrm{sin}\theta \approx \mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}=\frac{y}{D}$

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

$\frac{a∆{y}_{1,-1}}{D}=\left\{1-\left(-1\right)\right\}\lambda \phantom{\rule{0ex}{0ex}}D=\frac{a∆{y}_{1,-1}}{2\lambda }$

Substitute the values to get

$D=\frac{0.4mm×1.8mm}{2×450nm}\phantom{\rule{0ex}{0ex}}=\frac{0.4×{10}^{-3}m×1.8×{10}^{-3}m}{2×450×{10}^{-9}m}\phantom{\rule{0ex}{0ex}}=0.8m$

Thus, the distance is $0.8m$ .

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

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

$\frac{a∆{y}_{1,3}}{D}=\left\{3-1\right\}\lambda \phantom{\rule{0ex}{0ex}}∆{y}_{1,3}=\frac{2\lambda D}{a}\phantom{\rule{0ex}{0ex}}=\frac{2×450×{10}^{-9}m×0.8m}{0.4×{10}^{-3}m}\phantom{\rule{0ex}{0ex}}=1.8mm$

Thus, the distance is $1.8mm$ . ### Want to see more solutions like these? 