Interference in YDSE

Electricity and Magnetism Level 5

Let us take a YDSE (Young's Double Slit Experiment) set-up as shown:

If monochromatic source is emitting a light with power 'P'. Then find the position $x$ (as shown) of this source such that the contrast of the fringe pattern observe on screen is best possible.

Details and Assumptions

Contrast at any point on screen is defined as the difference in maximum and minimum intensity. If difference is more than contrast is also better, and we can easily detect the fringe pattern with our eyes.

x = \cfrac { d }{ 3\sqrt { 3 } }

x = \cfrac { d }{ 2\sqrt { 2 } }

x = \cfrac { d }{ \sqrt { 2 } }

x = d

1 solution

Deepanshu Gupta
Mar 22, 2015

Hints : (As requested by karan) Best contrast is possible when net Minimimum Intensity should be minimum to minimum (as possible) means it will be zero . which occures when intensity from both slits are equal.

Hence stablished relation b/w them and maximise the intensity and as far as I remembered final expression is

$\displaystyle{P\propto I\propto \cfrac { x }{ { ({ x }^{ 2 }+{ (\cfrac { d }{ 2 } ) }^{ 2 }) }^{ \cfrac { 3 }{ 2 } } } \\ \cfrac { dP }{ dx } =0\quad (\because \quad I\quad is\quad max)\\ x=\cfrac { d }{ 2\sqrt { 2 } } }$

@KARAN SHEKHAWAT I have posted hints , and I think this helps you .

Deepanshu Gupta - 6 years, 2 months ago

@Spandan Senapati bro can you post a solution? Thnx

Harsh Shrivastava - 3 years, 8 months ago

So the intensity of the light received by each slit is $P _{r}=Acos\theta I=PA\frac {x}{4\pi (x^2+d^2/4)^{3/2}}$ with a path difference of zero..The diagram has to be a bit more clear,coz I mistook it as the incident power is $P/4\pi r^2$ . The area is to be projected for getting the energy due to Normal incidence.

Spandan Senapati - 3 years, 8 months ago

But the sharpness of the fringes keeps on increasing as we go on increasing $x$

Spandan Senapati - 3 years, 8 months ago

Interference in YDSE

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