Q150 / 180 · COMEDK 2019 · Physics
In a Young's double slit experiment, the separation between the two slits is $0.9 \mathrm{~mm}$ and the fringes are observed $1 \mathrm{~m}$ away. If it produces the second dark fringe at a distance of $1 \mathrm{~mm}$ from the central fringe, the wavelength of the monochromatic source of light used is
- A$500 \mathrm{~nm}$
- Correct answerB$600 \mathrm{~nm}$
- C$450 \mathrm{~nm}$
- D$400 \mathrm{~nm}$
Solution
Given, separation between two slits,
$$
d=0.9 \mathrm{~mm}=0.9 \times 10^{-3} \mathrm{~m}
$$
Distance between slits and screen, $D=1 \mathrm{~m}$
The distance between two consecutive dark or bright fringes is given as $\beta$ (fringe width) and that between central tringe and first dark fringe on either side is $\frac{\beta}{2}$.
It is given that the spacing between second dark fringe and central fringe $=\beta+\frac{\beta}{2}=\frac{3 \beta}{2}$
As we know, $\beta=\frac{\lambda L}{d}$
Hence, $\frac{3}{2} \times \frac{\lambda D}{d}=1 \times 10^{-3} \mathrm{~m}$
$$
\begin{aligned}
&\Rightarrow \frac{3}{2} \times \lambda \times \frac{1}{0.9 \times 10^{-3}}=1 \times 10^{-3} \mathrm{~m} \\
&\Rightarrow \lambda=\frac{2}{3} \times 10^{-3} \times \frac{0.9 \times 10^{-3}}{1}=0.6 \times 10^{-6} \mathrm{~m}
\end{aligned}
$$
$\therefore$ Wavelength of monochromatic source of light,
$$
\lambda=600 \times 10^{-9} \mathrm{~m}=600 \mathrm{~nm}
$$
Hence, the correct answer is option: 'B'.