Physics - Wave Optics Question with Solution | TestHub

Let the initial width of both slits be $w$. The intensity of light passing through each slit is directly proportional to its width. Let the initial intensity of each slit be $I$.

The maximum intensity in the interference pattern is given by:
$I_{max}=(\sqrt{I}+\sqrt{I}{)}^2=4I$

The minimum intensity is given by:
$I_{min}=(\sqrt{I}-\sqrt{I}{)}^2=0$

When one slit is made twice as wide as the other, its width becomes $2w$. The intensity of light from this slit becomes $2I$, while the intensity from the other slit remains $I$.

The new maximum intensity becomes:
$I_{max}^{\prime }=(\sqrt{2I}+\sqrt{I}{)}^2=I(\sqrt{2}+1{)}^2\approx 5.83I$

The new minimum intensity becomes:
$I_{min}^{\prime }=(\sqrt{2I}-\sqrt{I}{)}^2=I(\sqrt{2}-1{)}^2\approx 0.17I$

Comparing the initial and final intensities, we observe that $I_{max}^{\prime }>I_{max}$ and $I_{min}^{\prime }>I_{min}$.

Therefore, the intensities of both the maxima and the minima increase.

Answer: the intensities of both the maxima and the minima increase.