Question
Question: In a Youngs\(^,\)s double slit experiment , the intensity at the central maximum is \({I_0}\) The in...
In a Youngs,s double slit experiment , the intensity at the central maximum is I0 The intensity at a distance 2β above the central maximum is (where β = fringe width).
Solution
According to given question we have to find out the path difference and phase at a distance of 2β unit from central maxima .After finding these values we have to apply maximum intensity equation to find two solvable equation. After solving we get the value and finally we get the answer of the question.
Complete step by step solution:
Here,
Central maxima intensity is I0
As we already knows that β=dDλ where β=fringe width
D= separation between slits and screen
λ=wavelength of light used
d=separation between two slits
And also y=2Dλd where y is difference from center to maxima exceeds by 2β unit
And other conventions are same as above
Now, we know that path difference is equal to:
Δx=Dyd where Δx=path difference
Now putting value of y in path difference equation, and after solving
Δx=2Dλd×dD Δx=2y
We have Φ=λ2π×Δx where Φ=phase difference
After putting value of Δx in phase difference equation:
Φ=λ2π×2λ
Φ=π
Now apply maximum intensity equation:
Imax=I1+I2+2I1I2cosΦ
Where I1 and I2 are intensity from two slits
Now for central bright fringe
Let I1=I2=Is and
We also know that for central bright fringe
Hence ,
On further solving,
Io=4Is
Now apply maximum intensity equation for fringe above to central bright fringe
So, I=Is+Is+2IsIscosπ
On further solving I=0
Hence, we can say that intensity of light at a distance 2β unit above central maxima is 0.
Note: In young's double slit experiment we use two parallel slits and screen to get the fringes pattern.In these types of questions we talked about the phase difference of light waves and the path difference of light waves. And a maximum intensity equation to get the intensity at a particular distance from the central maxima.