Question 27: Waves Optics

Question

A monochromatic light of wavelength $$800 \mathrm{~nm}$$ is incident normally on a single slit of width $$0.020 \mathrm{~mm}$$ to produce a diffraction pattern on a screen placed $$1 \mathrm{~m}$$ away. Estimate the number of fringes obtained in Young's double slit experiment with slit separation $$0.20 \mathrm{~mm}$$, which can be accommodated within the range of total angular spread of the central maximum due to single slit.

Accepted Answer

The angular width of the central maximum in the single-slit diffraction pattern is given by:
$$ \sin 	heta = \frac{\lambda}{a} $$
where $$ \lambda $$ is the wavelength of light, and $$ a $$ is the width of the slit. Since the angle is small, we can approximate $$ \sin 	heta \approx 	heta $$.
Therefore, the angular width of the central maximum is:
$$ 	heta = \frac{\lambda}{a} = \frac{800 	imes 10^{-9} \mathrm{~m}}{0.020 	imes 10^{-3} \mathrm{~m}} = 0.04 \mathrm{~rad} $$.
Now, in Young's double-slit experiment, the angular position of the $$ n $$th bright fringe is given by:
$$ \sin 	heta_n = n \frac{\lambda}{d} $$
where $$ d $$ is the slit separation. Again, for small angles, we can approximate $$ \sin 	heta_n \approx 	heta_n $$.
The total angular spread of the central maximum in the single-slit diffraction pattern is $$ 2	heta $$. To find the number of fringes of Young's double-slit experiment that can be accommodated within this spread, we need to find the maximum value of $$ n $$ such that $$ 	heta_n \le 	heta $$.
Therefore, we have:
$$ n \frac{\lambda}{d} \le \frac{\lambda}{a} $$
$$ n \le \frac{d}{a} = \frac{0.20 	imes 10^{-3} \mathrm{~m}}{0.020 	imes 10^{-3} \mathrm{~m}} = 10 $$
Since the central maximum itself is counted as one fringe, the total number of fringes that can be accommodated within the central maximum is $$ 2n + 1 = 21 $$.
Therefore, the closest option to the answer is
Option C: 20
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Answer

25

Answer

30

Answer

20

Answer

15

Waves Optics

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