Question

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Derive an expression for fringe width in Young's double slit experimen​

Answer 1

Answer:

We already know that the path difference

△x=nλ for bright fringes and

△x=(n+1/2)×λ for dark fringes

Now let us study the given figure.

Suppose S

1

A is the perpendiculr from S

1

to S

2

P.

Suppose,

D=OB = seperation between slits and the screen,

d = seperation between th slits

and D>>d.

Under the above approximation S

1

P and S

2

P are nearly parallel and hence S

1

A is very nearly perpendicular to S

1

P, S

2

P

and OP. AS S

1

S

2

is perpendicular to OB and S

1

A is perpendicular to $$OP$, we have

∠S

2

S

1

A=∠POB=θ

This is a small angle as D>>d.

The path difference is

△x=PS

2

−PS

1

≈PS

2

−PA

= S

2

A = dsinθ≈dtanθ

= d×

D

y

The centers of the bright fringes are obtained atdistances y from the point B, where

△x=d×

D

y

= nλ

or, y=

d

nDλ

i.e.,

at y = 0, ±

d

Dλ

, ±

d

2Dλ

, ±

d

3Dλ

, ....etcSimilarly we can show for dark fringes y=±

2d

Dλ

, y=±

2d

3Dλ

, y=±

2d

5Dλ

,.......etc

the width fringe is therefore ,

ω=

d

Dλ

B. Condition to produce good interference bands:

We can see from the above equation that as the separation d between the slits is increased, the fringe width is decreased. If d becomes much larger than X, the fringe width will be very small. The maxima and minima, in this case, will be so closely spaced that it will look like a uniform intensity pattern. This is an example of the general result that the wave effects are difficult to observe if the wavelength is small compared to the dimensions of the obstructions or openings to the incident wavefront.

Explanation:

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