CBSE Class 12 · Wave OpticsLong Answer

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(a) In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is

9 : 25

. Find the ratio of the widths of the two slits.

Solution

(a) At a point

P

(distance

x

from centre

O

) the path difference between the waves from

S_{1}

and

S_{2}

Δ = \frac{x d}{D}

Bright fringe

(constructive):

Δ = nλ \Rightarrow x_{n} = \frac{nλ D}{d}

Dark fringe

(destructive):

Δ = (2 n - 1) \frac{λ}{2} \Rightarrow x_{n} = \frac{( 2 n - 1 ) λ D}{2 d}

.
The fringe width is the spacing between consecutive bright (or dark) fringes:

β = x_{n + 1} - x_{n} = \frac{( n + 1 ) λ D}{d} - \frac{nλ D}{d} = \frac{λ D}{d}

(b) Given

\frac{I _{min}}{I _{ma x}} = \frac{9}{25}

\frac{I _{1} - I _{2}}{I _{1} + I _{2}} = \frac{9}{25} = \frac{3}{5}

5 (I_{1} - I_{2}) = 3 (I_{1} + I_{2}) \Rightarrow 2 I_{1} = 8 I_{2} \Rightarrow \frac{I _{1}}{I _{2}} = 16

Since intensity (hence slit width) are proportional,

\frac{W _{1}}{W _{2}} = \frac{I _{1}}{I _{2}}

\frac{W _{1}}{W _{2}} = 16 : 1

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CBSE Class 12 · Wave OpticsLong Answer

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