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Question

In YDSE monochromatic light of wavelength600nm incident of slits as shown in figure.

If S1S2=3mm, OP=11mm then

  1. If α=0.36π degree then destructive interfaces at point P.
  2. If α=0.36π degree then constructive interfaces at point O.
  3. If α=0 then constructive interfaces at O
  4. Fringe width depends on α.

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Solution

Solution:

Step 1: Given information

Wavelength =600nm

S1S2=3mm

OP=11mm

Step 2: Calculation for part (a)

If α=0.36π degree then destructive interfaces at point P.

Δx=3×10-3×0.36π×π180+3×11×10-61Δx=3900

Now,

3900=(2n1)λ2n=7dest

Step 3: Calculation for part (b)

If α=0.36π degree then constructive interfaces at point O.

Δx=3mm0.36ππ180Δx=600nm

600nm=n×600nm

n=1const

Step 4: Calculation for part (c)

If α=0 then constructive interfaces at O.

Consider, α=0,Δx=0

Therefore, constructive interference:

Δx=dsinα+dsinθ

x=dα+dyD

Step 5: Calculation for part (d)

Fringe width depends on α.

Therefore, Fringe width does not depend on alpha, a::b::c .


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