Optics Ques 127
- In a Young’s double slit experiment, slits are separated by $0.5$ $mm$ and the screen is placed $150$ $ cm$ away. A beam of light consisting of two wavelengths, $650 $ $nm$ and $520$ $ nm$, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide, is
(a) $7.8 $ $mm$
(b) $9.75 $ $mm$
(c) $15.6 $ $mm$
(d) $1.56 $ $mm$
(2017 Main)
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Answer:
Correct Answer: 127.(a)
Solution:
Formula:
- Let $n$th bright fringes coincides, then
$\begin{aligned} y _{n _1} & =y _{n _2} \ \Rightarrow \quad & \frac{n _1 \lambda _1 D}{d}=\frac{n _2 \lambda _2 D}{d} \Rightarrow \frac{n _1}{n _2}=\frac{\lambda _1}{\lambda _2}=\frac{520}{650}=\frac{4}{5}\end{aligned}$
Hence, distance from the central maxima is
$ y=\frac{n _1 \lambda _1 D}{d}=\frac{4 \times 650 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}}=7.8 $ $mm $
BETA
