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figure shows three equidistant slits ill...

figure shows three equidistant slits illuminated by a monochromatic parallel beam of light. Let `BP_0 -AP_0 = lambda/3` and `Dgtgtlambda`.

(a) Show that d = `sqrt ((2lambdaD)//3)`
(b) Show that the intensity at `P_0` is three times the intensity due to any of the three slits
individually.

Text Solution

Verified by Experts

The correct Answer is:
A, B
(a) Given, ` BP_0 - AP_0 = lambda//3`
sqrt(D^2 + d^2) -D = lambda/3`
or ` sqrt (D^2+d^2) = (D+ (lambda/3))`
Squaring both sides, we get
` D^2 + d^2 = D^2 + lambda^2/9 + (2Dlambda)/3`
Since `lambdaltltD`, we can ignore the term `lambda_2/9`. By ignoring this term, we get the desired result.
`d= sqrt(2lambdaD)/3`
(b) Given, `BP_0 - AP_0 = Deltax_(12) = lambda/3`
`:. Delta phi_(12)` or `phi_(12) = 120^@ `
Now, `CP_0 - AP_0 = Deltax_(13) = sqrt(2d)^2 + d^2)- D`
` = D[1+((2d)/D)^2]^(1/2)- D`
`= D[ 1+ 1/2 xx (4d^2)/D^2] - D`
`= (2d^2)/D`
Substituting, `d=sqrt(2lambdaD)/3` or `d^2 = (2lambdaD)/3`
We get, `Deltax_(13) = 4/3 lambda`
`:. Delta phi(13)` or `phi(13) = 4/3(360^@) = 480^@`
` = 480^@ - 360^@ = 120^@`
Now, we know that in case of coherent sources amplitudes are first added by vector method.
So, let individual amplitude is `A_0`.


The resultant amplitude will be given by
` A = sqrt (A_(0)^2 + (2A_0)^2 + 2(A_0)(2A_0)cos 120^@)`
`sqrt 3 A_0`
`I prop A^2`
and amplitude has become `sqrt3` times, therefore resultant intensity will become 3 times.
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