Question

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900\stackrel{o}{A}$ was $1.2mm$ for parallel narrow slits placed $2mm$ apart. In this arrangement, if the salt separation is increased by one-and-half times the previous value, then the fringe width is:

• A. $0.9mm$
• B. $0.8mm$
• C. $1.8mm$
• D. $1.6mm$

Answer 1

The correct option is B $0.8mm$

Fringe width $\beta =\frac{\lambda D}{d}$, where all the symbols have their usual meaning.

$\therefore$ $\beta \propto \frac{1}{d}$

$⟹$ $\frac{{\beta }^{\prime }}{\beta }=\frac{d}{d^{\prime }}$ ...........(1)

Given : $d=2mm$ $\beta =1.2mm$

$\therefore$ $d^{\prime }=d+\frac{d}{2}=\frac{3}{2}d$

Using (1), $\frac{{\beta }^{\prime }}{1.2}=\frac{2}{3}$ $⟹{\beta }^{\prime }=0.8mm$