In a Young's double-slit experiment, light of wavelength $600 ext{ nm}$ is used to obtain interference fringes on a screen kept $1.4 ext{ m}$ away. If the separation between the two slits is $0.28 ext{ mm}$, calculate: The fringe width (distance between two consecutive bright fringes). The distance of the 4th bright fringe from the central maximum. The distance of the 4th dark fringe from the central maximum.

Q: In a Young's double-slit experiment, light of wavelength $600\text{ nm}$ is used to obtain interference fringes on a screen kept $1.4\text{ m}$ away. If the separation between the two slits is $0.28\text{ mm}$, calculate: The fringe width (distance between two consecutive bright fringes). The distance of the 4th bright fringe from the central maximum. The distance of the 4th dark fringe from the central maximum.

Given Data: Wavelength of light, $\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m} = 6 \times 10^{-7}\text{ m}$ Distance of screen from the slits, $D = 1.4\text{ m}$ Separation between the slits, $d = 0.28\text{ mm} = 0.28 \times 10^{-3}\text{ m} = 2.8 \times 10^{-4}\text{ m}$ --- Calculation of Fringe Width ($\beta$): Formula for fringe width is: $$\beta = \frac{\lambda D}{d}$| Substituting the given values: $$\beta = \frac{6 \times 10^{-7} \times 1.4}{2.8 \times 10^{-4}}$$ Simplifying the expression: $$\beta = \frac{8.4 \times 10^{-7}}{2.8 \times 10^{-4}}$$ $$\beta = 3 \times 10^{-3}\text{ m} = 3\text{ mm}$$ Answer 1: The fringe width is $3\text{ mm}$ (or $0.003\text{ m}$). --- Distance of the 4th Bright Fringe ($x4$):- Formula for the position of the $n$-th bright fringe from the central maximum is: $$xn = \frac{n \lambda D}{d} = n \beta$$ For the 4th bright fringe ($n = 4$): $$x4 = 4 \times \beta$$ $$x4 = 4 \times 3\text{ mm} = 12\text{ mm} = 1.2 \times 10^{-2}\text{ m}$$ Answer 2: The distance of the 4th bright fringe from the central maximum is $12\text{ mm}$ (or $0.012\text{ m}$). --- Distance of the 4th Dark Fringe ($x'4$):- Formula for the position of the $n$-th dark fringe from the central maximum is: $$x'n = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} = \left(n - \frac{1}{2}\right) \beta$$ For the 4th dark fringe ($n = 4$): $$x'4 = \left(4 - \frac{1}{2}\right) \beta$$ $$x'4 = 3.5 \times 3\text{ mm}$$ $$x'4 = 10.5\text{ mm} = 1.05 \times 10^{-2}\text{ m}$$ Answer 3: The distance of the 4th dark fringe from the central maximum is $10.5\text{ mm}$ (or $0.0105\text{ m}$).

Question

In a Young's double-slit experiment, light of wavelength $600	ext{ nm}$ is used to obtain interference fringes on a screen kept $1.4	ext{ m}$ away. If the separation between the two slits is $0.28	ext{ mm}$, calculate: The fringe width (distance between two consecutive bright fringes). The distance of the 4th bright fringe from the central maximum. The distance of the 4th dark fringe from the central maximum.

Accepted Answer

Given Data: Wavelength of light, $\lambda = 600	ext{ nm} = 600 	imes 10^{-9}	ext{ m} = 6 	imes 10^{-7}	ext{ m}$ Distance of screen from the slits, $D = 1.4	ext{ m}$ Separation between the slits, $d = 0.28	ext{ mm} = 0.28 	imes 10^{-3}	ext{ m} = 2.8 	imes 10^{-4}	ext{ m}$ --- Calculation of Fringe Width ($\beta$): Formula for fringe width is: $$\beta = \frac{\lambda D}{d}$| Substituting the given values: $$\beta = \frac{6 	imes 10^{-7} 	imes 1.4}{2.8 	imes 10^{-4}}$$ Simplifying the expression: $$\beta = \frac{8.4 	imes 10^{-7}}{2.8 	imes 10^{-4}}$$ $$\beta = 3 	imes 10^{-3}	ext{ m} = 3	ext{ mm}$$ Answer 1: The fringe width is $3	ext{ mm}$ (or $0.003	ext{ m}$). --- Distance of the 4th Bright Fringe ($x4$):- Formula for the position of the $n$-th bright fringe from the central maximum is: $$xn = \frac{n \lambda D}{d} = n \beta$$ For the 4th bright fringe ($n = 4$): $$x4 = 4 	imes \beta$$ $$x4 = 4 	imes 3	ext{ mm} = 12	ext{ mm} = 1.2 	imes 10^{-2}	ext{ m}$$ Answer 2: The distance of the 4th bright fringe from the central maximum is $12	ext{ mm}$ (or $0.012	ext{ m}$). --- Distance of the 4th Dark Fringe ($x'4$):- Formula for the position of the $n$-th dark fringe from the central maximum is: $$x'n = \left(n - \frac{1}{2}ight) \frac{\lambda D}{d} = \left(n - \frac{1}{2}ight) \beta$$ For the 4th dark fringe ($n = 4$): $$x'4 = \left(4 - \frac{1}{2}ight) \beta$$ $$x'4 = 3.5 	imes 3	ext{ mm}$$ $$x'4 = 10.5	ext{ mm} = 1.05 	imes 10^{-2}	ext{ m}$$ Answer 3: The distance of the 4th dark fringe from the central maximum is $10.5	ext{ mm}$ (or $0.0105	ext{ m}$).

Given Data:

1. Calculation of Fringe Width ($\beta$):

2. Distance of the 4th Bright Fringe ($x_4$):-

3. Distance of the 4th Dark Fringe ($x'_4$):-