Q129 / 180 · NEET 2013 (All India) · Physics
In Young's double slit experiment, the slits are $2 \mathrm{~mm}$ apart and are illuminated by photons of two wavelengths $\lambda_1=12000 Å \quad$ and $\lambda_2=10000 Å$. At what minimum distance from the common central bright fringe on the screen $2 \mathrm{~m}$ from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?
- A$8 \mathrm{~mm}$
- Correct answerB$6 \mathrm{~mm}$
- C$4 \mathrm{~mm}$
- D$3 \mathrm{~mm}$
Solution
Given $\lambda_1=12000 Å$ and $\lambda_2=10000 Å$, $D=2 \mathrm{~cm}$ and $d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~cm}$.
We have $\frac{\lambda_1}{\lambda_2}=\frac{h_2}{h_1}$
$$
=\frac{12000}{10000}=\frac{6}{5}
$$
as $x=\frac{u_1 \lambda_1 D}{d}=\frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}}$
$=5 \times 1.2 \times 10^4 \times 10^{-10} \times 10^3=6 \mathrm{~mm}$
Hence, the correct answer is option: 'B'.