Question 99: Wave Optics

Question

In Young's double slits experiment, the position of 5$$\mathrm{^{th}}$$ bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :

Accepted Answer

In Young's double-slit experiment, the distance between the slits and the screen, $L = 1 	ext{ m}$, the wavelength of the light, $\lambda = 600 	ext{ nm} = 600 	imes 10^{-9} 	ext{ m}$, and the position of the 5th bright fringe from the central maximum, $y = 5 	ext{ cm} = 0.05 	ext{ m}$.
The distance between the central maximum and the $n$th bright fringe is given by the formula:
$$y_n = \frac{n\lambda L}{d}$$
where $d$ is the distance between the two slits.
We can rearrange this formula to solve for $d$:
$$d = \frac{n\lambda L}{y_n}$$
Substituting the values given in the question, we get:
$$d = \frac{5 	imes 600 	imes 10^{-9} 	imes 1 	imes 100}{5}$$
$$d = 6 	imes 10^{-5} 	ext{ m}$$
$$d = 60 \ \mu	ext{m}$$
Therefore, the separation between the slits is $60 \ \mu	ext{m}$.

Answer

60 $$\mu$$m

Answer

48 $$\mu$$m

Answer

36 $$\mu$$m

Answer

12 $$\mu$$m

Wave Optics

In Young's double slits experiment, the position of 5$\mathrm{^{th}}$ bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :

Choose the correct answer: