Question

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900\mathring{A}$ was $1.2mm$ for parallel narrow slits placed $2mm$ 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 ${\beta }_1=1.2mm$
$\frac{d_2}{d_1}=1\frac{1}{2}=1.5$
So, $\beta \propto \frac{1}{d}$
$\frac{{\beta }_1}{{\beta }_2}=\frac{1/d_1}{1/d_2}$
$\frac{{\beta }_1}{{\beta }_2}=\frac{d_2}{d_1}=1.5$
$\Rightarrow \frac{1.2}{{\beta }_2}=1.5$
$\Rightarrow {\beta }_2=\frac{1.2}{1.5}=\frac{4}{5}=0.8mm$