Question 1: Waves Optics

Question

Distance of 5th dark fringe from centre is $$4 \mathrm{~mm}$$. If $$D=2 \mathrm{~m}, \lambda=600 \mathrm{~nm}$$, then distance between slits is

Accepted Answer

Given, distance between source and observer,
$$D=2 \mathrm{~m}$$
Distance of 5th dark fringe from centre $$=4 	imes 10^{-3} \mathrm{~m}$$
Wavelength, $$\lambda=600 \mathrm{~nm}=6 	imes 10^{-7} \mathrm{~m}$$
Distance of 5th dark fringe from centre
$$=\beta+\beta+\beta+\beta+\frac{\beta}{2}=\frac{9 \beta}{2}$$
where, $$\beta=$$ fringe width
According to question,
$$\begin{aligned}
& \frac{9 \beta}{2}=4 	imes 10^{-3} \
& \frac{9}{2} 	imes \frac{D \lambda}{d}=4 	imes 10^{-3} \
& \frac{9}{2} 	imes \frac{2 	imes 6 	imes 10^{-7}}{d}=4 	imes 10^{-3} \
& \Rightarrow \quad d=1.35 	imes 10^{-3} \mathrm{~m}=1.35 \mathrm{~mm}
\end{aligned}$$

Answer

$$1.35 \mathrm{~mm}$$

Answer

$$2.00 \mathrm{~mm}$$

Answer

$$3.25 \mathrm{~mm}$$

Answer

$$10.35 \mathrm{~mm}$$

Waves Optics

Distance of 5th dark fringe from centre is $4 \mathrm{~mm}. If D=2 \mathrm{~m}, \lambda=600 \mathrm{~nm}$, then distance between slits is

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