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Asked by swayamagarwal2114 17th July 2022, 12:05 PM
Answered by Expert
Answer:
Part (i)
 
Light ray incident on slits S1 and S2 at angle α = (1.8/π)o .
 
Hence additional phase Δφ2 at S2 = ( d sinα ) × ( 2π / λ ) 
 
additional phase Δφ2 at S2 calculated from above expression is
 
Δφ2 = 10-3 ×  sin (1.8/π) × [ 2π / ( 6000× 10-10 ) ] = (100/3)π
 
After slit S1 , there is a transparent film of thickness t and refractive index μ .
 
hence phase difference Δφ1 between light rays emerging from S1 and S2 is
 
Δφ1 = ( μ - 1 ) t × ( 2π / λ )  = (1/2) × 4 × 10-6 × [ 2π / ( 0.6 × 10-6 ) ]
 
Δφ1 = (20/3)π
 
Net phase difference = Δφ2 - Δφ1 = (80/3)π
 
Let Io be the intensity of light emerging from slit S1 . Thin film absorbs 50% intensity.
Hence after thin film light intensity ( Io / 2 )
 
Width of slit S2 is twice of that of S1 . hence light intensity from slit S2 is ( 2 Io )
 
when two light rays of intensities I1 and I2 are added , we get resultant intensity IR as
 
IR = I1 + I2 + 2 begin mathsize 14px style square root of I subscript 1 I subscript 2 end root end style cosΔφ 
where Δφ is the phase difference between two light rays.
---------------------------------
 
Part (i)
 
Intensity at centre point of screen = (Io/2) + 2Io  + 2 [ (Io/2)(2Io) ]1/2 cos [ (80/3)π ) = (3/2) Io
 
Maximum intensity occurs at a point where phase difference is zero
 
Hence Maximum intensity = (Io/2) + 2Io  + 2 [ (Io/2)(2Io) ]1/2  = (9/2) Io
 
If we denote maximum intensity (9/2) I= I ,
 
then Intensity at centre point of screen = { [ (3/2) Io ] / [ (9/2) Io ] } × I =  (1/3) I
 
-------------------------
Part (ii)
 
Minimum intensity is when phase difference Δφ = π
 
Mimimum intensity = (Io/2) + 2Io  + 2 [ (Io/2)(2Io) ]1/2 cos [ π ] = Io/2
 
If we denote maximum intensity (9/2) I= I ,
 
then Intensity at centre point of screen = { [ (1/2) Io ] / [ (9/2) Io ] } × I =  (1/9) I 
 
----------------------------------------
Part (iii)
 
Maxima of intensity occure at angular position θn , if following condition is satisfied
 
Δp + ( d sinθn ) = nλ
 
where Δp is initial phase difference at slits position.
 
Let yn be the vertical distance from O on screen for n-th maximum, then we have
 
Δp + ( d × (yn/D ) = nλ  .......................... (1)
 
Let yn+1 be the vertical distance from O on screen for (n+1)-th maximum, then we have

Δp + ( d × (yn+1/D ) = (n+1)λ  ......................(2)
 
By subracting eqn.(1) from eqn.(2) , we get
 
(d/D) [ yn+1 - yn ] = λ
 
Hence fringe width = [ yn+1 - yn ] = ( λ D ) /d  = ( 6000 × 10-10 ) / 10-3  = 6 × 10-4 m
 
fringe width = 0.6 mm
 
-----------------------------------------------------
Part (iv)

we have seen from Part (i) ,Net phase difference at O = Δφ2 - Δφ1 = (80/3)π
 
Net phase difference at O = [13+ (1/3)]2π
 
To get phase difference as integer multiples of 2π ,
 
required additional phase difference =  (2/3) (2π )  = (4/3)π
 
Hence required path difference = (4/3)π × ( λ / 2π ) = (2/3) λ
 
angular position θ to get this additional phase difference is determined from the following equation
 
d sinθ ≈ d tanθ = d × (y/D) = (2/3)λ
 
where y is distance of nearest maximum from O and D is slit-to-screen distance
 
y = [ (2/3) × 6000 × 10-10 ] / ( 10-3 ) = 4 × 10-4 m = .4 mm
 
Hence nearest maximum from O is at adistance y = 0.4 mm
-------------------------------
part (v)
 
Central maximum occurs when phase difference between two rays emerging from slits S1 and S2 is zero.
 
we have seen from Part (i) ,Net phase difference at O = Δφ2 - Δφ1 = (80/3)π
 
Equivalenet path difference = (80/3)π × ( λ / 2π ) = (40/3) λ 
 
To reduce above path difference , required angular position θ is determined from the following expression
 
d sinbegin mathsize 14px style theta end style begin mathsize 14px style almost equal to end style d tanbegin mathsize 14px style theta end style = (40/3) λ
d tanbegin mathsize 14px style theta end style = d ( y / D )  = (40/3) λ
y = (40/3d) λ = [ 40 / ( 3 × 10-3 ) ] ( 6000 × 10-10 ) = 8 × 10-3 m = 8 mm
 
Hence central maximum occurs at a point 8 mm above O
 
-------------------------------------------------------------------
part ( vi )
 
Path difference at a distance 4 mm above O is determined as
 
Δp = d sinθ ≈ d tanθ = 10-3 × (4/1000) = 4 × 10-6 m
 
Equivalent phase difference = 4 × 10-6 × ( 2π / λ ) = 4 × 10-6 × [ 2π / ( 6000 × 10-10 ) ]

Equivalent phase difference =(40/3)π
 
Phase didfference at a distance 4 mm above O  = (80/3)π - (40/3)π = (40/3)π
 
IR = I1 + I2 + 2 begin mathsize 14px style square root of I subscript 1 I subscript 2 end root end style cosΔφ
(Io/2) + 2Io  + 2 [ (Io/2)(2Io) ]1/2 cos [ (40/3)π ) = (3/2) Io
 
If we denote maximum intensity (9/2) I= I ,
 
then Intensity at 4 mm above O = { [ (3/2) Io ] / [ (9/2) Io ] } × I =  (1/3) I
 
-----------------------------------------------------------------------------------------------
Answered by Expert 17th July 2022, 11:49 PM
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