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# YDSE

Asked by swayamagarwal2114 17th July 2022, 12:05 PM
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 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

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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

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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
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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 sin d tan = (40/3) λ
d tan = 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

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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 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

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Answered by Expert 17th July 2022, 11:49 PM
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