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  • In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is <img alt="\mathrm{5 \mathrm{~mm}}" src="https://entrancecorner.oncodec

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is \mathrm{5 \mathrm{~mm}}. The screen on which the diffraction pattern is displayed is at a distance of \mathrm{80 \mathrm{~cm}} from the slit. The wavelength is \mathrm{6000 \, \AA}. The slit width (in mm) is about:

Option: 1

0.576


Option: 2

0.348


Option: 3

0.192


Option: 4

0.096


Answers (1)

best_answer

Distance between the first minimum on the left and the first minimum on the right is also the width of central maximum.
Width of central maximum, \mathrm{W=\frac{2 \lambda D}{a}}
Where, \mathrm{\lambda=} Wavelength of light

\mathrm{\mathrm{a}=}  Width of the slit

\mathrm{\mathrm{D}=} Distance of the screen from the slit

\mathrm{ \therefore \mathrm{a}=\frac{2 \lambda \mathrm{D}}{\mathrm{W}} }

Here, \mathrm{\lambda=6000 \AA=6000 \times 10^{-10} \mathrm{~m}}

\mathrm{ \begin{aligned} & \mathrm{w}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m} \\\\ & \therefore \mathrm{a}=\frac{2 \times 6000 \times 10^{-10} \mathrm{~m} \times 80 \times 10^{-2} \mathrm{~m}}{5 \times 10^{-3} \mathrm{~m}} \\\\ & =19.2 \times 10^{-5} \mathrm{~m}=0.192 \times 10^{-3} \mathrm{~m}=0.192 \mathrm{~mm} \end{aligned} }

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