Question

The fringe width in Young's double slit experiment is $2\times {10}^{-4}m$ if the distance between the slits is halved and the slit screen distance is double, then the new fringe width will be :

• A. $2\times {10}^{-4}m$
• B. $1\times {10}^{-4}m$
• C. $0.5\times {10}^{-4}m$
• D. $8\times {10}^{-4}m$

Answer 1

The correct option is D $8\times {10}^{-4}m$

We know that,

fringe width$\left(\beta \right)=\frac{\lambda D}{d}$

where,

$\lambda \to$ wavelength of light

$D\to$ distance between slit and screen

$d\to$ distance between slits

So,

we have,

$\frac{\lambda D}{d}=2\times {10}^{-4}$ ……..(i) (given)

Now,

If the distance between slits is halved and slit screen distance is double, then the new finge width $\left({\beta }_o\right)$

${\beta }_o=\frac{\lambda \left(2D\right)}{d/2)}=\frac{4\lambda D}{d}$

$\Rightarrow {\beta }_o=\frac{4\lambda D}{d}$ …….(ii)

From (i)

${\beta }_o=4(2\times {10}^{-4})$

${\beta }_o=8\times {10}^{-4}$m.