Question

In a double slit experiment the distance between slits is increased ten times whereas their distance from screen is halved then the fringe width is :

• A. Becomes $\frac{1}{20}$
• B. Becomes $\frac{1}{90}$
• C. It remains same
• D. Becomes$\frac{1}{10}$

Answer 1

Answer: (A)

Becomes $\frac{1}{20}$

Answer 2

: Fringe width

$\beta = \frac{D\lambda}{d}$

According to the question

$D' = \frac{D}{2}andd' = 10d$

$\therefore\beta' = \frac{D'\lambda}{d'} = \frac{(D/2)\lambda}{10d} = \frac{1}{20}\frac{D\lambda}{d}$

$\Rightarrow \beta' = \frac{\beta}{20}$ (using (i))