0
$\begingroup$

I had a doubt on finding the ratio of the width of the slits in the Young's Double-slit experiment.

The ratio of the intensities at minima to the maxima is already given as $9:25$. Looking for a solution for this problem, I've found a formula: $$\frac{I_{max}}{I_{min}}=\frac{{(a_1+a_2)}^2} {{(a_1-a_2)}^2}$$ Plugging in my values, I got the answer as $16:1$

So my question is, is there a general proof to this formula? I'm sure I've not learned this in my classes so an explanation of this formula is deeply appreciated.

$\endgroup$
0
$\begingroup$

$a_1$ and $a_2$ are the amplitudes of the waves arriving from two slits and creating an interference pattern.

For constructive interference the waves arrive in phase and so the amplitudes of the resultant wave is $a_1+a_2$ and as intensity is proportional to the amplitude squared you arrive at the maximum intensity $I_{\rm max} \propto (a_2+a_1)^2$.
For destructive interference you arrive at $I_{\rm min} \propto (a_2-a_1)^2$ and assuming that the constant of proportionality is the same in both case gives your quoted equation.

Your answers of $16\,:\, 1$ is the ratio

$\text{intensity of light from slit 1 alone} \, : \, \text{intensity of light from slit 2 alone}$

in the region where the interference occurred which might well have beeen due to the slits being of unequal width.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.