Q124 / 160 · TS EAMCET 2014 · Physics
In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 Å$ was $1.2 \mathrm{~mm}$ for parallel narrow slits placed $2 \mathrm{~mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is
- A$0.9 \mathrm{~mm}$
- Correct answerB$0.8 \mathrm{~mm}$
- C$1.8 \mathrm{~mm}$
- D$1.6 \mathrm{~mm}$
Solution
By Young's double slit interference experiment
$$
\beta=\frac{\lambda D}{d}
$$
The given
$$
\begin{aligned}
& \beta_1=1.2 \mathrm{~mm} \\
& \frac{d_2}{d_1}=1 \frac{1}{2}=1.5
\end{aligned}
$$
So, $\quad \beta \propto \frac{1}{d}$
$$
\begin{aligned}
& \frac{\beta_1}{\beta_2}=\frac{1 / d_1}{1 / d_2} \\
& \frac{\beta_1}{\beta_2}=\frac{d_2}{d_1}=1.5 \Rightarrow \frac{1.2}{\beta_2}=1.5 \\
\Rightarrow \quad & \beta_2=\frac{1.2}{1.5}=\frac{4}{5}=0.8 \mathrm{~mm}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.