In a double slit experiment, I_o is the intensity of the central bright fringe obtained with monochromatic light of $λ = 6000Å$. Determine the intensity at a distance of $4.8×10^{−5}m$ from the central maximum if the separation between the slits is 0.25cm and the distance between the screen and the double slit is 1.20m.

$0.75I_o$

From the principle of superposition the resultant intensity $I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos δ$,

where $δ$ = phase difference = $\frac{2π}{λ}$ × path difference.

At the central fringe, $δ = 0$, thus $I_0 = I_1+I_2+2I_1 = 4I_1$

At a distance x from the central bright fringe, the path difference = $\frac{xd}{D}$ and the corresponding phase difference $δ =\frac{2π}{λ}\frac{xd}{D}$

Hence the resultant intensity at the assigned position will be

$I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos δ =\frac{I_o}{4}+\frac{I_o}{4}+\frac{2I_o}{4}\cos δ$

$\frac{I_o}{4}[1+1+2\cos δ]=\frac{I_o}{2}[1+\cos δ]=I_o \cos^2(\frac{δ}{2})$

Here, $\frac{δ}{2}=\frac{1}{2}\frac{2π}{λ}\frac{xd}{D}=\frac{π}{6}$

Thus $I=\frac{I_o}{2}×(\frac{\sqrt{3}}{2})^2=\frac{3I_o}{4}=0.75I_o$