Question 41: Waves Optics

Question

In Young's double slit experiment, the two slits are separated by 0.2 mm and they are 1 m from the screen. The wavelength of the light used is 500 nm. The distance between 6th maxima and 10th minima on the screen is closest to

Accepted Answer

Given, $$d=0.2 \mathrm{~mm}=2 	imes 10^{-4} \mathrm{~m}$$
$$D=1 \mathrm{~m}, \lambda=500 \mathrm{~nm}=5 	imes 10^{-7} \mathrm{~m}$$
The distance between 6th maxima and 10th minima is given as
$$\begin{aligned}\Delta x & =\left(x_{10}ight)_{	ext {dark }}-\left(x_6ight)_{	ext {bright }} \
& =\frac{(2 	imes 10-1) D \lambda}{2 d}-\frac{6 D \lambda}{d} \
& =\frac{D \lambda}{d}\left[\frac{19}{2}-6ight) \
& =\frac{1 	imes 5 	imes 10^{-7}}{2 	imes 10^{-4}}\left[\frac{7}{2}ight]=\frac{35}{4} 	imes 10^{-3} \mathrm{~m} \
& =8.75 	imes 10^{-3} \mathrm{~m} \
& =8.75 \mathrm{~mm}
\end{aligned}$$
Hence, $$8 \mathrm{~mm}$$ is closest to $$8 \mathrm{~mm}$$.

Answer

12 mm

Answer

10 mm

Answer

14 mm

Answer

8 mm

Waves Optics

In Young's double slit experiment, the two slits are separated by 0.2 mm and they are 1 m from the screen. The wavelength of the light used is 500 nm. The distance between 6th maxima and 10th minima on the screen is closest to

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