Question

In a double slit experiment, the distance between slits is increased $10$ times where as their distance from the screen is halved, then the fringe width?

• A. remains same
• B. Becomes $1/10$th
• C. Becomes $1/20$th
• D. Becomes $1/90$th

Answer 1

The correct option is C Becomes $1/20$th

Let $\lambda$ be wavelength of monochromatic light, $d$ the distance between coherent sources, and $D$ the distance between screen and source, then fringe width is
$\beta =\frac{D\lambda }{d}$
Given, $d_1=d,D_1=D,d_2=10d,D_2=\frac{D}{2}$
$\therefore$ ${\beta }_2=\frac{\frac{D}{2}\lambda }{10d}=\frac{D\lambda }{20d}$
$\Rightarrow$ ${\beta }_2=\frac{W}{20}$