Question 28: Waves Optics

Question

Incident light of wavelength $$\lambda=800 \mathrm{~nm}$$ produces a diffraction pattern on a screen $$1.5 \mathrm{~m}$$ away when it passes through a single slit of width $$0.5 \mathrm{~mm}$$. The distance between the first dark fringes on either side of the central bright fringe is

Accepted Answer

To find the distance between the first dark fringes on either side of the central bright fringe in a single-slit diffraction pattern, we need to use the formula for the position of the dark fringes. The position of the first dark fringe is given by:
$$ y = \frac{m \lambda L}{a} $$
where:
$$m$$ is the order of the dark fringe ($$m = 1$$ for the first dark fringe)
$$\lambda$$ is the wavelength of the light
$$L$$ is the distance from the slit to the screen
$$a$$ is the width of the slit
Given:
$$\lambda = 800 \ \mathrm{nm} = 800 	imes 10^{-9} \ \mathrm{m}$$
$$L = 1.5 \ \mathrm{m}$$
$$a = 0.5 \ \mathrm{mm} = 0.5 	imes 10^{-3} \ \mathrm{m}$$
For the first dark fringe ($$m = 1$$):
$$ y_1 = \frac{1 	imes 800 	imes 10^{-9} \ \mathrm{m} 	imes 1.5 \ \mathrm{m}}{0.5 	imes 10^{-3} \ \mathrm{m}} $$
Simplifying:
$$ y_1 = \frac{800 	imes 1.5 	imes 10^{-9+3}}{0.5} \ \mathrm{m} $$
$$ y_1 = \frac{1200 	imes 10^{-6}}{0.5} \ \mathrm{m} $$
$$ y_1 = \frac{1200 	imes 10^{-6}}{0.5} \ \mathrm{m} $$
$$ y_1 = 2400 	imes 10^{-6} \ \mathrm{m} $$
$$ y_1 = 2.4 \ \mathrm{mm} $$
This is the distance from the central bright fringe to the first dark fringe on one side. Therefore, the total distance between the first dark fringes on either side of the central bright fringe is:
$$ 2 	imes 2.4 \ \mathrm{mm} = 4.8 \ \mathrm{mm} $$
Thus, the correct answer is:
Option D: 4.8 mm

Answer

2.4 mm

Answer

2.4 cm

Answer

4.8 cm

Answer

4.8 mm

Waves Optics

Incident light of wavelength $\lambda=800 \mathrm{~nm} produces a diffraction pattern on a screen 1.5 \mathrm{~m} away when it passes through a single slit of width 0.5 \mathrm{~mm}$. The distance between the first dark fringes on either side of the central bright fringe is

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