Question

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900\, ??$ 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)

0.8 mm

Answer 2

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, $\beta \propto \frac{1}{d}$ $\frac{\beta_{1}}{\beta_{2}}=\frac{1 / d_{1}}{1 / d_{2}}$ $\frac{\beta_{1}}{\beta_{2}}=\frac{d_{2}}{d_{1}}=1.5$ $\Rightarrow \frac{1.2}{\beta_{2}}=1.5$ $\Rightarrow \beta_{2}=\frac{1.2}{1.5}=\frac{4}{5}=0.8 \,mm$