Question 344: Wave Optics

Question

In Young's double slit experiment, the fringe width is $2 \mathrm{~mm}$. The separation between the $13^{	ext {th }}$ bright fringe and the $4^{	ext {th }}$ dark fringe from the centre of the screen on same side will be

Accepted Answer

Given: Fringe width $\mathrm{W}=2 \mathrm{~mm}$
The distance of the $\mathrm{n}^{	ext {th }}$ bright fringe from centre of the screen $y_n=\frac{n \lambda D}{d}$ ...... (i)
The distance of the $\mathrm{n}^{	ext {th }}$ dark fringe from centre of the screen $y_n^{\prime}=(2 n-1) \frac{\lambda D}{2 d}$ ..... (ii)
Substituting, $\mathrm{n}=13$ in (i) and $\mathrm{n}=4$ in (ii) we get $\mathrm{y}_{13}=\frac{13 \lambda \mathrm{D}}{\mathrm{d}}$ and $\mathrm{y}_4^{\prime}=\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}$
$	herefore \quad$ The separation between the $13^{	ext {th }}$ bright fringe and the $7^{	ext {th }}$ dark fringe is
$\begin{aligned}
& \mathrm{y}_{13}-\mathrm{y}_4^{\prime} \
& =\frac{13 \lambda \mathrm{D}}{\mathrm{d}}-\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}} \
& =\left(13-\frac{7}{2}ight) \frac{\lambda \mathrm{D}}{\mathrm{d}} \
& =\frac{19}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{19}{2} \mathrm{~W} \
& 	ext { but } \mathrm{W}=2 \mathrm{~mm} \
	herefore \quad & \mathrm{y}_{13}-\mathrm{y}_4^{\prime}=\frac{19}{2} 	imes 2=19 \mathrm{~mm}
\end{aligned}$

Answer

$13 \mathrm{~mm}$.

Answer

$17 \mathrm{~mm}$.

Answer

$19 \mathrm{~mm}$.

Answer

$23 \mathrm{~mm}$.

Wave Optics

In Young's double slit experiment, the fringe width is 2 \mathrm{~mm}. The separation between the 13^{\text{th }} bright fringe and the 4^{\text{th }} dark fringe from the centre of the screen on same side will be

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