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  • The maximum intensity in Young's double slit experiment is Distance between the slits is <i

The maximum intensity in Young's double slit experiment is \mathrm{I_0.} Distance between the slits is \mathrm{d= 5 \lambda,} where \mathrm{\lambda} is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance \mathrm{D=10 \mathrm{~d}}

Option: 1

\mathrm{\frac{I_0}{2}}


Option: 2

\mathrm{\frac{3}{4} \mathrm{I}_0}


Option: 3

\mathrm{I_0}


Option: 4

\mathrm{\frac{\mathrm{I}_0}{4}}


Answers (1)

best_answer

Suppose P is a point infront of one slit at which intensity is to be calculated from figure it is clear that \mathrm{x=\frac{d}{2}}. Path difference between the waves reaching at P

\mathrm{ \Delta=\frac{x d}{D}=\frac{\left(\frac{d}{2}\right) d}{10 d}=\frac{d}{20}=\frac{5 \lambda}{20}=\frac{\lambda}{4} }

Hence corresponding phase difference \mathrm{\varphi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2} }

Resultant intensity at P
\mathrm{ I=I_{\max } \cos ^2 \frac{\varphi}{2}=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2} }

Posted by

Ritika Jonwal

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