Optics Ques 134
- 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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Answer:
Correct Answer: 134.(d)
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 $ $mm$
BETA
