Question 398: Waves Optics

Question

Two slits in Young's double slit experiment are $$1.5 \mathrm{~mm}$$ apart and the screen is placed at a distance of $$1 \mathrm{~m}$$ from the slits. If the wavelength of light used is $$600 	imes 10^{-9} \mathrm{~m}$$ then the fringe separation is

Accepted Answer

To find the fringe separation (also known as fringe width) in Young's double slit experiment, we use the formula:
$$ \beta = \dfrac{\lambda D}{d} $$
where:
$$\beta$$ is the fringe width (fringe separation).
$$\lambda$$ is the wavelength of the light used.
$$D$$ is the distance from the slits to the screen.
$$d$$ is the distance between the two slits.
Given values are:
$$d = 1.5 \, \mathrm{mm} = 1.5 	imes 10^{-3} \, \mathrm{m}$$
$$D = 1 \, \mathrm{m}$$
$$\lambda = 600 	imes 10^{-9} \, \mathrm{m}$$
Substituting these values into the formula:
$$ \beta = \dfrac{600 	imes 10^{-9} \, \mathrm{m} 	imes 1 \, \mathrm{m}}{1.5 	imes 10^{-3} \, \mathrm{m}} $$
Calculating the value:
$$ \beta = \dfrac{600 	imes 10^{-9}}{1.5 	imes 10^{-3}} $$
$$ \beta = 4 	imes 10^{-4} \, \mathrm{m} $$
Thus, the fringe separation is $$4 	imes 10^{-4} \, \mathrm{m}$$. The correct answer is:
Option D $$4 	imes 10^{-4} \mathrm{~m}$$

Answer

$$4 	imes 10^{-5} \mathrm{~m}$$

Answer

$$9 	imes 10^{-8} \mathrm{~m}$$

Answer

$$4 	imes 10^{-7} \mathrm{~m}$$

Answer

$$4 	imes 10^{-4} \mathrm{~m}$$

Waves Optics

Two slits in Young's double slit experiment are $1.5 \mathrm{~mm} apart and the screen is placed at a distance of 1 \mathrm{~m} from the slits. If the wavelength of light used is 600 \times 10^{-9} \mathrm{~m}$ then the fringe separation is

Choose the correct answer: