Physics - Wave Optics Question with Solution | TestHub

Let the intensity of light from the two slits be $I_1=I$ and $I_2=2I$.

The maximum intensity in the interference pattern is given by:
$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2=(\sqrt{I}+\sqrt{2I}{)}^2=I(1+\sqrt{2}{)}^2=I(3+2\sqrt{2})$

The minimum intensity is given by:
$I_{min}=(\sqrt{I_2}-\sqrt{I_1}{)}^2=(\sqrt{2I}-\sqrt{I}{)}^2=I(\sqrt{2}-1{)}^2=I(3-2\sqrt{2})$

The ratio of maximum to minimum intensity is:
$\frac{I_{max}}{I_{min}}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

Rationalizing the denominator:
$\frac{I_{max}}{I_{min}}=\frac{(3+2\sqrt{2}{)}^2}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{17+12\sqrt{2}}{9-8}=17+12\sqrt{2}$

Substituting $\sqrt{2}\approx 1.414$:
$\frac{I_{max}}{I_{min}}\approx 17+12(1.414)=17+16.968=33.968$

Rounding to the nearest integer, the ratio is approximately $34:1$.

Answer: $34:1$