2016 Wave OpticsYoung's double slit experiment – intensity at a point Medium Question 25 of 40 in Wave Optics

The intensity at the maximum in a Young's double slit experiment is $I_0$. Distance between two slits is $d = 5\lambda$, where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity infront of one of the slits on the screen placed at a distance D = 10 d?

A $\frac{I_0}{4}$
B $\frac{3}{4}I_0$
C $\frac{I_0}{2}$
D $I_0$

The answer is C.

Solution

Path difference λ/4 gives phase π/2 and intensity $I_0/2$.

Intensity $I = I_0\cos^2\left(\frac{\phi}{2}\right)$ ...(i), where $\phi = \frac{2\pi}{\lambda}\Delta X$ ...(ii)

In front of one slit, $y = \frac{d}{2}$, so $\Delta X = d\frac{y}{D} = \frac{d\times\frac{d}{2}}{10d} = \frac{d}{20}$
With $d = 5\lambda$: $\Delta X = \frac{5\lambda}{20} = \frac{\lambda}{4}$ ...(iii)
$\phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{4} = \frac{\pi}{2}$
$I = I_0\cos^2\left(\frac{\pi}{4}\right) = I_0\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$

Concept ydse-intensity-in-front-of-slit

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40 questions from this chapter have appeared across 18 exam years.