The intensity at the maximum in a Young's double slit experiment is I0 . Distance between two slits is d=5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D=10 d?

From the double slit experiment we get to know that

Path difference = S2P – S1P

\(\begin{array}{l}sqrt{(D^{2}+d^{2})-D}\end{array} \)

\(\begin{array}{l}D(1+frac{1}{2}frac{d^{2}}{D^{2}}-1)\end{array} \)

\(\begin{array}{l}frac{d^{2}}{2D}\end{array} \)

=

\(\begin{array}{l}Delta x=frac{d^{2}}{2times 10d}\end{array} \)

\(\begin{array}{l}frac{d}{20}\end{array} \)

=

\(\begin{array}{l}frac{5lambda }{20}\end{array} \)

=

\(\begin{array}{l}frac{lambda }{4}\end{array} \)

=

\(\begin{array}{l}Delta phi =frac{2Pi }{lambda } times frac{lambda }{4}=frac{pi }{2}\end{array} \)

The intensity at the desired point will be

I=I0cos2(ϕ/2)=I0cos2(π/4)=I0/2

Answer

I0/2

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