0
$\begingroup$

Here's a question I just came around
In YDSE, the intensity of the maxima is I.If the width of each slit is doubled, what will be the intensity of maxima now ?
here, we assume that no diffraction is occurring.

what I thought was that the intensity is power per unit area, therefore even though more light is coming in but the area factor will cancel it hence the intensity must remain same.

But the answer was 2I and was according to the direct relation of slit width and the intensity of light.

please tell where I m wrong?

Also, I saw a proof online where the person is treating two dependent variables as independent.

Here's the link check it too.

$\endgroup$
4
  • $\begingroup$ @Farcher but the OP says "no diffraction is occurring", which means geometric optics, which means they're not slits, they're big holes, which mean the intensity doesn't change. $\endgroup$
    – JEB
    Commented Sep 29, 2019 at 15:27
  • $\begingroup$ @JED I take your point. $\endgroup$
    – Farcher
    Commented Sep 29, 2019 at 15:44
  • $\begingroup$ @JED I want to clarify we are just assuming that diffraction is not happening, although it will happen in reality... $\endgroup$ Commented May 20, 2020 at 13:42
  • $\begingroup$ @Farcher I want to clarify we are just assuming that diffraction is not happening, although it will happen in reality... $\endgroup$ Commented May 20, 2020 at 13:42

1 Answer 1

1
$\begingroup$

The width of the slits has nothing to do with the separation of the fringes.

Doubling the width of each of the two slits leaves the separation of the fringes the same but each fringe will now have twice a much light forming it.

So the answer is $2I$.

$\endgroup$
3
  • $\begingroup$ No,as I said in my question the intensity must not change..please tell where I am in that logic $\endgroup$ Commented May 20, 2020 at 15:24
  • $\begingroup$ Also another case which might be considered is when we take power of source to be same(quite convincing) then as width doubles area becomes twice therefore intensity will decrease.... $\endgroup$ Commented May 20, 2020 at 15:26
  • $\begingroup$ In any case ,Intensity must not increase $\endgroup$ Commented May 20, 2020 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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