Physics - Wave Optics Question with Solution | TestHub

The fringe width in Young's double slit experiment is given by:

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

The change in fringe width when the screen is moved by a distance $\Delta D$ is:

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

Substituting the given values:

$3\times 10^{-5}=\frac{600\times 10^{-9}\times \Delta D}{10^{-3}}$

$3\times 10^{-5}=6\times 10^{-4}\times \Delta D$

$\Delta D=\frac{3\times 10^{-5}}{6\times 10^{-4}}=0.05\text{ m}=5\text{ cm}$

Since the fringe width can be changed (either increased or decreased) by $3\times 10^{-5}\text{ m}$, the screen can be moved away from the slits by $5\text{ cm}$ or moved towards the slits by $5\text{ cm}$.

Thus, both (a) and (b) are correct.

Answer: both (a) and (b)