Q10.5    In Young’s double-slit experiment using monochromatic light of wavelength \lambda , the intensity of light at a point on the screen where path difference is \lambda , is K units. What is the intensity of light at a point where path difference is \frac{\lambda}{3}?

Answers (1)

Given, in youngs double-slit experiment.

the wavelength of monochromatic light = \lambda

The intensity of light when the path difference is \lambda = K

Now,

As we know,

The phase difference \phi is given by

   \phi =\frac{2\pi }{\lambda }(PathDiffernce)

also 

Total Intensity

 I=I_1+I_2+2\sqrt{I_1I_2}cos\phi  

Let  I_1=I_2=I_0

Now, when path difference is \lambda

the phase difference angle 

 \phi=\frac{2\pi }{\lambda}*\lambda=2\pi

so,

I_0+I_0+2\sqrt{I_0I_0}cos2\pi=K

K=4I_0

 

Now, when path difference is    \frac{\lambda }{3}

\phi=\frac{2\pi }{\lambda}*\frac{\lambda}{3}=\frac{2\pi}{3}

Intensity of light

 K'=I_0+I_0+2\sqrt{I_0I_0}cos\frac{2\pi}{3}

K'=I_0

Now comparing intensity at both cases

\frac{K'}{K}=\frac{I_0}{4I_0}=\frac{1}{4}

K'=\frac{K}{4}

Hence intensity will reduce to one-fourth of initial when path difference changes from \lambda to    \frac{\lambda}{3}.

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