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Minimum distance from S2 on x-axis where intensity of light is maximum

2 votes

S1 and S2 are two coherent sources placed along the y axis with a S2 at origin distance between S2 and S1 is 4$\lambda $ then minimum distance from S2 on x-axis where intensity of light is maximum.
here's the diagram given:

what I did first thought ydse but answer comes out to be wrong. as I don't know about the screen .
then what Shoud I do?

asked Feb 23, 2017 in Physics Problems by physicsapproval (2,320 points)
Is the answer is 7.5$\lambda$
no its 1.17 $\lambda $

1 Answer

3 votes
Best answer

Their will be maximum intensity when there it is maxima .
For maxima path difference should be integral multiple of wavelength.
There are nine points where it will be maxima .
That is when $\Delta x=0,\lambda , 2\lambda ,3\lambda $and$ 4\lambda $ and similarly for negative x axis .

Now path difference is 4$\lambda $ when x=0 . When it is 0 then x= infinity .

So the point where it is maxima nearest to S$_2$ is when path difference is $3\lambda$.
Let the point be at a distance a from S$_2$ .

Then we get the equation as
$\sqrt{(4\lambda)^2 + a^2}- a=3\lambda$

Solving it we get a=1.17$\lambda$(approx)

answered Feb 23, 2017 by koolman (4,286 points)
edited Feb 25, 2017 by koolman
what's a and I don't understand how you got that equation, can you draw figure?
'a' is x coordinate . Let that point be P where it is maximum intensity .
Then I simply write the equation as
$S_1 P-S_2P =\Delta x$
find $S_1 P $ by pythagorus.
how you arrived  at the above equation is not clear to me
Which equation
$S_1 P-S_2P =\Delta x$
$\sqrt{(4\lambda)^2 + a^2}- a=3\lambda$
Sorry but I need diagram
thanks alot axtually I was not understanding why you subtracted a again as I was taking $\delta $x wrong .