QJEEAdv13Ph148c: JEE Advanced 2013 Physics - Class 12 JEEAdv13Ph148c Solution: In the Young's double-slit experiment using a monochromatic…

Wavelength of light is $\lambda$ .

Now, using the relation for intensity, we have

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

${1 \over 2} = {\cos ^2}\left( {{\phi \over 2}} \right) \Rightarrow \cos {\phi \over 2} = {1 \over {\sqrt 2 }}$

$\Rightarrow {\phi \over 2} = {\pi \over 4} \Rightarrow \phi = {\pi \over 2}$

$\Rightarrow \phi = {\pi \over 2},{{3\pi } \over 2},{{5\pi } \over 2},{{7\pi } \over 2}$

$\phi = {{2\pi } \over \lambda }\Delta x \Rightarrow {\pi \over 2} = {{2\pi } \over 2}\Delta x \Rightarrow \Delta x = {\lambda \over 4},{{3\lambda } \over 4},{{5\lambda } \over 4}$

$\Rightarrow \Delta x = (2n + 1){\lambda \over 4}$