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Maxima & Minima in YDSE

In a YDSE bic...

Question

In a YDSE bichromatic light of wavelength $400 n m$ and $560 n m$ are used. The distance between the slits is $0.1 m m$ and the distance between the plane of the slits and the screen is $1 m$ . The minimum distance between two successive regions of complete darkness is

A

4 m m

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B

5.6 m m

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C

14 m m

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D

28 m m

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Solution

The correct option is D $28 m m$
$\begin{matrix} L e t n t h min i m a o f 400 n m c o i n c i d e s w i t h m t h min i m a o f 560 n m t h e n, (2 n - 1) 400 = (2 m - 1) 560 \Rightarrow \frac{2 n - 1}{2 m - 1} = \frac{7}{5} = \frac{14}{10} = \frac{21}{15} i . e . 4 t h min i m a o f 400 n m c o i n c i d e s w i t h 3 r d min i m a o f 560 n m . = \frac{7 (1000) (400 \times 10^{- 6})}{2 \times 0.1} = 14 m m n e x t, 11 t h min i m a o f 400 n m w i l l c o i n c i d e w i t h 8 t h min i m a o f 560 n m l o c a t i o n o f t h i s min i m a i s = \frac{21 (1000) (400 \times 10^{- 6})}{2 \times 0.1} = 42 m m] ∴ r e q u i r e d d i s tan c e = 28 m m \end{matrix}$
Hence,
option $(D)$ is correct answer.

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