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Q102P

Expert-verifiedFound in: Page 1115

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Monochromatic light (wavelength${\mathbf{=}}{\mathbf{450}}{\mathbf{}}{\mathit{n}}{\mathit{m}}$ ) is incident perpendicularly on a single slit (width${\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{4}}{\mathbf{}}{\mathit{m}}{\mathit{m}}$ ). 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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{8}}{\mathit{m}}{\mathit{m}}$ . **

**(a) What is the distance from the slit to the screen? ( Hint: The angle to either minimum is small enough that${\mathit{s}}{\mathit{i}}{\mathit{n}}{\mathit{\theta}}{\mathbf{\approx}}{\mathit{t}}{\mathit{a}}{\mathit{n}}{\mathit{\theta}}$ .) **

**(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 $1.8mm$ .

Slit width $a=0.4mm$

Distance between the two minima on either side of the central maxima localid="1663048702619" $\u2206{y}_{1,-1}=1.8mm$

Wavelength of incident light $\lambda =450nm$

**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

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

Here, $\lambda $ is the wavelength of the incident light.

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 $m\pm 1$ on either sides of the central maxima is

$\frac{a\u2206{y}_{1,-1}}{D}=\left\{1-\left(-1\right)\right\}\lambda \phantom{\rule{0ex}{0ex}}D=\frac{a\u2206{y}_{1,-1}}{2\lambda}$

Substitute the values to get

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

Thus, the distance is $0.8m$ .

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

$\frac{a\u2206{y}_{1,3}}{D}=\left\{3-1\right\}\lambda \phantom{\rule{0ex}{0ex}}\u2206{y}_{1,3}=\frac{2\lambda D}{a}\phantom{\rule{0ex}{0ex}}=\frac{2\times 450\times {10}^{-9}m\times 0.8m}{0.4\times {10}^{-3}m}\phantom{\rule{0ex}{0ex}}=1.8mm$

Thus, the distance is $1.8mm$ .

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