Question

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

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

Answer 1

Answer: (A)

We know that, fring width
$\beta =\frac{D \lambda }{d}$ ....(i)
When distance slits is increased 10 times and distance between between the source and screen is halved, then new fringe width,
$\beta ^{′}=\frac{D^{′} \lambda }{d ′}$
$= \frac{D \lambda }{2 \times 10 \times d}$
$=\frac{1}{20}\cdot \frac{D \lambda }{d}$
$=\frac{\beta }{20}$ [from Equation (i)]