Optics 6 Question 19
21. In a YDSE bi-chromatic light of wavelengths $400 nm$ and $560 nm$ are used. The distance between the slits is $0.1 mm$ 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
(2004, 2M)
(a) $4 mm$
(b) $5.6 mm$
(c) $14 mm$
(d) $28 mm$
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Solution:
- Let $n$th minima of $400 nm$ coincides with $m$ th minima of $560 nm$, then
$$ \begin{aligned} (2 n-1) \frac{400}{2} & =(2 m-1) \frac{560}{2} \\ \text { or } \quad \frac{2 n-1}{2 m-1} & =\frac{7}{5}=\frac{14}{10}=\ldots \end{aligned} $$
i.e. 4th minima of $400 nm$ coincides with $3 rd$ minima of $560 nm$.
Location of this minima is,
$$ \begin{aligned} Y _1 & =\frac{(2 \times 4-1)(1000)\left(400 \times 10^{-6}\right)}{2 \times 0.1} \\ & =14 mm \end{aligned} $$
Next 11th minima of $400 nm$ will coincide with 8th minima of $560 nm$.
Location of this minima is,
$$ Y _2=\frac{(2 \times 11-1)(1000)\left(400 \times 10^{-6}\right)}{2 \times 0.1}=42 mm $$
$\therefore \quad$ Required distance $=Y _2-Y _1=28 mm$
