In Young's double-slit experiment, the point source $S$ is placed slighttly off the central axis as shown in figure. If $λ = 500 n m$ , then match the following.

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Solution

Due to elevation of source from the central axis, the path difference between rays reaching sources $S_{1}$ is $\frac{2 m m \times 20 m m}{1 m} = 4 \times 10^{- 5} m$ .
For interference at point P, path difference at that point= $\frac{x d}{D} + 4 \times 10^{- 5}$
$= \frac{10 m m \times 20 m m}{2 m} + 4 \times 10^{- 5}$
$= 1.4 \times 10^{- 4} m = 280 λ$ .
For interference at point O,
path difference at the point= $4 \times 10^{- 5} m = 80 λ$
When the transparent paper is introduced, the introduced path difference= $t (μ - 1) = 9 \times 10^{- 6} = 18 λ$
Hence at point P, the new order of bright fringes can be $298 λ$ or $262 λ$ depending upon which o the two slits is covered.
At point O, the new order of bright fringes can be $98 λ$ or $62 λ$ depending upon which o the two slits is covered.

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Similar questions

λ = 500 n m, d = 10 m m a n d D = 1.0 m

S

λ = 500 nm

Column $(I)$	Column $(II)$
$(A)$ Nature and order of interference at point $P$ , $O P = 10 mm$	$1.$ Bright fringe of order $80$
$(B)$ Nature and order of interference at point $O$	$2.$ Bright fringe of order $262$
$(C)$ If a transparent paper (refractive index $μ = 1.45$ ) of thickness $t = 0.02 mm$ is pasted on $S_{1}$ , i.e., one of the slits, the nature and order of interference at point $P$	$3.$ Bright fringe of order $62$
$(D)$ After inserting the transparent paper in front of Slit $S_{1}$ , the nature and order of interference at $O$ .	$4.$ Bright fringe of order $280$

P

λ = 500 nm

I_{0}

λ = 500 n m, d = 1 m m

D = 1 m

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