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In a double slit experiment, \mathrm{I}_{0}, is the intensity of the central bright fringe obtained with monochromatic light of \lambda=6000 \mathrm{~A}^{\circ}. Determine the intensity at a distance of 4.8 \times 10^{-5} \mathrm{~m} from the central maximum if the separation between the slits is 0.25 \mathrm{~cm} and the distance between the screen and the double slit is 1.20 \mathrm{~m}.

Option: 1

0.25 \mathrm{I}_{\mathrm{o}}


Option: 2

0.50 \mathrm{I}_{\mathrm{o}}


Option: 3

0.75 \mathrm{I}_{\mathrm{o}}


Option: 4

\mathrm{I}_{\mathrm{o}}


Answers (1)

best_answer

From the principle of superposition the resultant intensity I=I_{1}+I_{2}+2 \sqrt{ } I_{1} I_{2} \cos \delta,
Where \delta=  phase difference =\frac{2 \pi}{\lambda} \times \text{path difference }.
At the central fringe, \mathrm{\delta=0, thus \: I_{0}=I_{1}+I_{2}+2 I_{1}=4 I_{1}}

At a distance x from the central bright fringe, the path difference \mathrm{=\frac{x d}{D}}  and the corresponding phase difference \mathrm{\delta=\frac{2 \pi}{\lambda} \frac{\mathrm{xd}}{\mathrm{D}}}

Hence the resultant intensity at the assigned position will be
\mathrm{I=I_{1}+I_{2}+2 \sqrt{ } I_{1} I_{2} \cos \delta=\frac{I_{0}}{4}+\frac{I_{o}}{4}+\frac{2 I_{o}}{4} \cos \delta}
\mathrm{=\frac{I_{0}}{4}[1+1+2 \cos \delta]=\frac{I_{o}}{2}[1+\cos \delta]=I_{0} \cos ^{2}\left(\frac{\delta}{2}\right)}
\mathrm{Here, \frac{\delta}{2}=\frac{1}{2} \frac{2 \pi}{\lambda} \frac{x d}{D}=\frac{\pi}{6}}
\mathrm{Thus \mathrm{I}=\frac{\mathrm{I}_{\mathrm{o}}}{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3 \mathrm{I}_{\mathrm{o}}}{4}=0.75 \mathrm{I}_{\mathrm{o}}}

Posted by

Ritika Harsh

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