Maharashtra State PSC Model Paper 2

Concept: Let the intensity of light passing through the two slits be ${\mathrm{I}}_1$ and ${\mathrm{I}}_2$ . The intensity at a point having a phase difference δ is given by,
$\mathit{I}={\mathit{I}}_1+{\mathit{I}}_2+2\sqrt{{\mathit{I}}_1}{\mathit{I}}_2\cos \mathit{\delta }$
The above equation gives the intensity at maxima $\left(\mathit{\delta }=0\right)$ and minima $\left(\mathit{\delta }=\mathit{\pi }\right)$ as,
${\mathit{I}}_{\max }={\mathit{I}}_1+{\mathit{I}}_2+2\sqrt{{\mathit{I}}_1}{\mathit{I}}_2$
${\mathit{I}}_{\min }={\mathit{I}}_{\mathit{l}}+{\mathit{I}}_2+2\sqrt{{\mathit{I}}_1}{\mathit{I}}_2$
When the slits are of the same width, ${\mathit{I}}_2={\mathit{I}}_0{\mathit{I}}_2={\mathit{I}}_0,$
${\mathit{I}}_{\max }=4{\mathit{I}}_0,{\mathit{I}}_{\min }=0$
${\mathit{I}}_{\max }=4{\mathit{I}}_0,{\mathit{I}}_{\min }=0$
When the width of one of the slit is doubled,
Now the maximum and minimum intensity will be
${\mathit{I}}_{\max }=\left(3+2\sqrt{2}\right){\mathit{I}}_0=5.83{\mathit{I}}_0$
${\mathit{I}}_{\min }=\left(3-2\sqrt{2}\right){\mathit{I}}_0=0.17{\mathit{I}}_0$
From the above values, it is clear that the intensities of both maxima and minima increase.