Optics 6 Question 12
14. 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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Solution:
- 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
$$ y=\frac{n _1 \lambda _1 D}{d}=\frac{4 \times 650 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}}=7.8 mm $$
