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In young’s double slit experiment, the point source S is placed slightly off the central axis as shown in figure. If $λ$ = 500 nm, then match the following
COLUM N-I COLUM N-II
A) Nature and order of interference at the point P, OP=10mm P) Bright fringe of order 80
B) Nature and order of interference point O Q) Bright fringe of order 262
C) If a transparent paper (refractive index $μ$ = 1.45 ) of thickness f = 0.02 mm is passed on S₁ i.e. one of the slits, the nature and order of interference at P R) Bright fringe of order 62
D) After inserting the transparent paper in front of slits S₁, the nature and order of interference at O S) Bright fringe of order 280
(A) (B) (C) (D)
1) P Q S R
2) R P S Q
3) S P Q R
4) S Q R P

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A-P, B-Q, C-S, D-R

A-R, B-P, C-S, D-Q

A-S, B-P, C-Q, D-R

A-S, B-Q, C-R, D-P

answer is C.

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Detailed Solution

The optical path difference the two waves arriving at P is

$\begin{array}{l} δ = ({SS}_{2} + S_{2} P) - ({SS}_{1} + S_{1} P) \\ = ({SS}_{2} - {SS}_{1}) + (S_{2} P - S_{1} P) \\ = dsin θ_{0} + dsin θ = \frac{{dy}_{0}}{D_{1}} + \frac{dy}{D_{2}} \\ d = 20 mm y_{0} = 2 mm D_{2} = 2 m, D_{1} = 1 m, y = 10 m \\ δ = \frac{20 \times 2}{1000} + \frac{20 \times 10}{2000} = 0.14 mm \end{array}$
For a bright fringe, $δ = \frac{{dy}_{0}}{D_{1}} = 0.04 mm$
$n^{'} = \frac{δ^{'}}{λ} = \frac{0.04}{0.5 \times 10^{- 3}} = 80$
Due to transparent paper, the change in optical path
$\begin{array}{l} = (μ - 1) t = (1.45 - 1) (0.02) mm = 0.009 mm \\ δ^{H} = 0.14 mm - 0.009 mm = 0.131 mm \end{array}$
$n = \frac{0.131}{0.5 \times 10^{- 3}} = 262$
Due to transparent paper, the path difference at O
$\begin{array}{l} δ^{H} = δ^{1} - (μ - 1) t = (0.04 - 0.009) mm = 0.0131 m \\ n = \frac{0.131}{0.5 \times 10^{- 3}} = 62 \end{array}$

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