Question

In a double slit interference experiment, the fringe width obtained with a light of wavelength $ 5900\, \mathring{A} $ was $ 1.2\,mm $ for parallel narrow slits placed $ 2\,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 mm
• B. 0.8 mm
• C. 1.8 mm
• D. 1.6 mm

Answer 1

Answer: (B)

By Young's double slit interference experiment
$\beta=\frac{\lambda D}{d}$
The given $\beta_{1}=1.2 \,mm$
$\frac{d_{2}}{d_{1}}=1 \frac{1}{2}=1.5$
So, $\beta \propto \frac{1}{d}$
$ \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 \beta_{2}=\frac{1.2}{1.5}=\frac{4}{5}=0.8 \,mm $