1
$\begingroup$

This is a question that came in my examination-

In a Young's double slit, one of the slits is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern"

Now, I suppose this could be interpreted two ways- either the ratio of the maximum intensity of the fringe with and without the glass, or that of the light and dark fringes.

If intensity is given by $$\\I= 4I'{{cos^2 \phi\\}}/{{2}}$$ then I don't see why there should be a difference in the ratio of the intensity of the bright and dark fringes, because both their intensities will have decreased.

If that is not the case, it must be the first part. Could someone help in how both the slits affect the intensity, quanititatively?

$\endgroup$
  • 1
    $\begingroup$ C.B.S.E :) :) :P $\endgroup$ – SmarthBansal Mar 7 '18 at 16:44
  • $\begingroup$ @SmarthBansal, I'm assuming you gave the exam too. Did you get the correct answer? $\endgroup$ – Antara Kulkarni Mar 9 '18 at 4:29
  • $\begingroup$ I think so. it was something like (3+2*2^(0.5))^2. You ready for chemistry? it's gonna be tough. $\endgroup$ – SmarthBansal Mar 9 '18 at 6:02
  • 1
    $\begingroup$ @SmarthBansal yeah... I'm kind of dreading chemistry $\endgroup$ – Antara Kulkarni Mar 12 '18 at 9:49
2
$\begingroup$

The resultant amplitude of two interfering waves is $$ {A_{net}}^2 = {A_1}^2 + {A_2}^2 +2A_1A_2\cos\theta $$ where $\theta$ is the phase difference between the waves.

Since intensity is proportional to the square root of the amplitude we have $$I_{net} = I_1 + I_2 +2\root\of{I_1}\root\of{I_2}\cos\theta$$ Normally in a double slit experiment the sources are same and coherent and that gives $$ I_1=I_2=I' (say)$$ and the formula for $I_{net}$ reduces to the one you mentioned.

But Since the source intensities are not same you can't apply that formula.
Instead just use $ I_1=\frac{I_2}{2}=I_0 $.
For the maximum intensity $\cos\theta = 1$ and for minimum intensity $\cos\theta = -1$

Find the $I_{net}$ for both these cases and take the ratio. You get
$$ (3+2\root\of2)^2$$

$\endgroup$
1
$\begingroup$

The intensity is not given by the equation by $\\I= 4I'{{cos^2 \phi\\}}/{{2}}$ as this equation is derived on the assumption that the intensity of the waves from both slits is the same.
This equation predicts that the intensity minimum will be zero.

If the amplitudes of the superposing waves from the two slits is not equal as would be the case if the light intensities from the two slits are not the same then the minimum will no longer be zero.
In such a case the minimum intensity will be proportional to $(\rm amplitude_{\text{slit 1}}- amplitude_{\text{slit 2}})^2$. The maximum intensity will also need to be adjusted.

$\endgroup$
  • $\begingroup$ I see, could you perhaps give a more definitive answer? $\endgroup$ – Antara Kulkarni Mar 9 '18 at 4:28
  • $\begingroup$ @AntaraKulkarni Use the intensity at the two slits to find the ratio of the amplitudes of the waves originating from each of the slits. $\endgroup$ – Farcher Mar 9 '18 at 5:56

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.