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Q.

In YDSE the separation between slits  S1 and S2   is 5mm and slits are illuminating with light of wavelength 5000A0 . The interference fringes are formed on the screen with central maximum at ‘O’. A glass plate of μ=32  with thickness 1μm  at end A and is linearly increasing to 7μm  at end B is moved in front of slits with constant speed  1mm/sec  downwards. At t=0, end A is right in front of S1  . AB=6mm

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a

At t=0, just after end A of plate comes in front of slit 1, the order of maxima at O is 1.

b

At t=5 sec 5th  bright fringe is at O

c

Order of maxima at O when B is in front of slit 1 is 5.

d

Order of maxima at O when B is in front of slit 1 is 7. 

answer is A, B, C.

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Detailed Solution

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y=αx+C       1μm=α×0+C         C=1μm            7μm=α×6mm+1μm

m = 103          S1S2=d=5mm,λ=5000A°             μ=1.5

(A)  (μ1)tλ=0.5×1×1065000×1010=1
(B) At t = 5s,   x=1mms×5s=5mm
  y=5μm+1μm=6μm
 No. of fringes crossed   0.5×6×1065000×1010=6
 6th bright fringe will be at O
(C,D) Here  x=6mm
y=6μm+1μm=7μm
 No. of fringes crossed =  0.5×7×1065000×1010=7
 

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