Question
In Young's double$-$slit experiment, the distance between two consecutive bright fringes on a screen placed at $1.5 m$ from the two slits is $0.6\ mm$. What would be the fringe width, if the screen is brought towards the slits by $50\ cm$, keeping rest of the setting the same?

Answer

Let $\lambda$ be the wavelengths of light used and $d$ the distance between the two sources $($i.e., slits$)$,
if $D$ is the distance between the sources and the screen, the fringe width is
$w =\frac{\lambda D}{d}$
For the same $\lambda$ and $d, W \propto D$.
$\therefore \frac{W_2}{W_1}=\frac{D_2}{D_1}$
Data: $W_1=0.6\ mm =6 \times 10^{-4} m ,$
$D_1=1.5 m , D_2=1 m$
$\therefore W_2=\frac{W_1 D_2}{D_1}$
$=\frac{6 \times 10^{-4} \times 1}{1.5}$
$= 4 \times 1 0 ^{- 4 }m~$
This is the required fringe width.

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