Question 2: Waves Optics

Question

A light of wavelength $$500 \mathrm{~nm}$$ is incident on a Young's double slit. The distance between slit and screen is $$D=1.8 \mathrm{~m}$$ and distance between slits is $$d=0.4 \mathrm{~mm}$$. If screen moves with a speed of $$4 \mathrm{~m} / \mathrm{s}$$, then with what speed first maxima will move?

Accepted Answer

Given, wavelength of light,
$$\lambda=500 \mathrm{~nm}=5 	imes 10^{-7} \mathrm{~m}$$
Distance between slit and screen, $$D=1.8 \mathrm{~m}$$
Distance between two slits,
$$d=0.4 \mathrm{~mm}=4 	imes 10^{-4} \mathrm{~m}$$
Velocity of screen $$=\frac{d D}{d t}=4 \mathrm{~m} / \mathrm{s}$$
Speed of first maxima $$=\frac{d \beta}{d t}=$$ ?
In Young's double slit experiment, fringe width, $$\beta=\frac{D \lambda}{d}$$
Differentiating with respect to $$t$$, we get
$$\begin{aligned}
\frac{d \beta}{d t} & =\frac{\lambda}{d} \cdot \frac{d D}{d t}=\frac{5 	imes 10^{-7}}{4 	imes 10^{-4}} 	imes 4 \
& =5 	imes 10^{-3} \mathrm{~m} / \mathrm{s}=5 \mathrm{~mm} / \mathrm{s}
\end{aligned}$$

Answer

5 mm/s

Answer

4 mm/s

Answer

3 mm/s

Answer

2 mm/s

Waves Optics

A light of wavelength $500 \mathrm{~nm} is incident on a Young's double slit. The distance between slit and screen is D=1.8 \mathrm{~m} and distance between slits is d=0.4 \mathrm{~mm}. If screen moves with a speed of 4 \mathrm{~m} / \mathrm{s}$, then with what speed first maxima will move?

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