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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