A Young's double-slit experiment is performed using monochromatic light of wavelength \lambda . The intensity of light at a point on the screen, where the path difference is \lambda , is K units. The intensity of light at a point where the path difference is \lambda \over 6 is given by nK \over 12 , where n is an integer. The value of n is \underline\hspace2cm .

Wave Optics PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q50 with Solution

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Correct answer: 9

In Young’s double-slit experiment, if the two slits have equal intensity, the resultant intensity at path difference $\Delta$ is

$I = I_{\max}\cos^2\left(\frac{\phi}{2}\right)$

where the phase difference

$\phi = \frac{2\pi \Delta}{\lambda}.$

Equivalently,

$I = I_{\max}\cos^2\left(\frac{\pi \Delta}{\lambda}\right).$

= I_{\max}\cos^2(\pi) = I_{\max}.$$ Given this intensity is $K$, so $$K = I_{\max}.$$ 3. At the point where path difference is $\Delta = \frac{\lambda}{6}$: $$I_2 = I_{\max}\cos^2\left(\frac{\pi}{6}\right) = I_{\max}\left(\frac{\sqrt{3}}{2}\right)^2 = I_{\max}\cdot \frac{3}{4}.$$ Since $I_{\max} = K$, $$I_2 = \frac{3K}{4}.$$ 4. According to the question, $$I_2 = \frac{nK}{12}.$$ So, $$\frac{nK}{12} = \frac{3K}{4}.$$ Cancelling $K$: $$\frac{n}{12} = \frac{3}{4}$$ $$n = 12\cdot \frac{3}{4} = 9.$$ 5. Therefore, $$\boxed{n=9}$$