Question

Youngs double slit interference zderivations of equation

Answer 1

YOUNG'S DOUBLE SLIT INTERFERENCE

DERIVATION :

In the above diagram,

the path difference is S2N

for small angle sinα = Tanα = α

therefore, in triangle S1S2N

$\alpha = \frac{s2n}{s1s2}$

from the diagram it clearly states that,

S1S2 = d , implies

$\alpha = \frac{s2n}{d}$

therefore,

$s2n \: = \alpha d$

where S2N is the path difference.

from PAO

α = PO / AO

α = X / D ---------(1 )

substituting (1) in path difference gives,

path diff = X . d / D

condition for bright fringe :
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path diff = n¥

here( ¥ denotes wavelength)

implies, Xd/D = n¥

X = n¥D / d ( BRIGHT FRINGE)

X1 = ¥D/ d

X2 = 2¥D/d

X2 -X1 = ¥D / d

β = ¥D / d ------(3)

condition for dark fringe :
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path diff = (2n-1)¥ / 2

implies, Xd / D = (2n - 1 ) ¥ / 2

therefore,

X = ( 2n - 1 ) ¥ D / 2d

when we substitute n = 1

then X 1= ¥D / 2d ,

for n= 2.

X 2 = 3¥D / d

β = X2 -X1 = ¥D / d -------( 3 )

therefore the fringe width for both bright and dark fridge will be same .

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Answer 2

Answer:

The two slits in Young's double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. No interference pattern will be observed on the screen. ... It will change so quickly that there will be general illuminated and hence interference pattern will no t be observed.