In Young's double-slit experiment, monochromatic light of wavelength $589 ext{ nm}$ is used. The distance between the two slits is $0.5 ext{ mm}$ and the screen is kept at a distance of $1.0 ext{ m}$ from the slits. Calculate: The fringe width (distance between two consecutive bright fringes). The angular fringe width in radians. What will be the change in fringe width if the entire apparatus is immersed in water of refractive index $4/3$?

Q: In Young's double-slit experiment, monochromatic light of wavelength $589\text{ nm}$ is used. The distance between the two slits is $0.5\text{ mm}$ and the screen is kept at a distance of $1.0\text{ m}$ from the slits. Calculate: The fringe width (distance between two consecutive bright fringes). The angular fringe width in radians. What will be the change in fringe width if the entire apparatus is immersed in water of refractive index $4/3$?

Given Data: Wavelength of light in air ($\lambda$) = $589\text{ nm} = 589 \times 10^{-9}\text{ m}$ Distance between slits ($d$) = $0.5\text{ mm} = 0.5 \times 10^{-3}\text{ m} = 5 \times 10^{-4}\text{ m}$ Distance of screen from slits ($D$) = $1.0\text{ m}$ Refractive index of water ($\mu$) = $4/3 = 1.33$ --- Part 1: Calculation of Fringe Width ($\beta$)\nThe formula for fringe width in Young's double-slit experiment is: $$\beta = \frac{\lambda D}{d}$$ \nSubstituting the given values: $$\beta = \frac{(589 \times 10^{-9}\text{ m}) \times (1.0\text{ m})}{5 \times 10^{-4}\text{ m}}$$ $$\beta = \frac{589 \times 10^{-9}}{5 \times 10^{-4}}$$ $$\beta = 117.8 \times 10^{-5}\text{ m} = 1.178 \times 10^{-3}\text{ m} = 1.178\text{ mm}$| --- Part 2: Calculation of Angular Fringe Width ($\theta$)\nThe formula for angular fringe width is: $$\theta = \frac{\beta}{D} = \frac{\lambda}{d}$| \nSubstituting the values: $$\theta = \frac{589 \times 10^{-9}\text{ m}}{5 \times 10^{-4}\text{ m}}$$ $$\theta = 117.8 \times 10^{-5}\text{ radians} = 1.178 \times 10^{-3}\text{ rad}$$ --- Part 3: Change in Fringe Width in Water\nWhen the apparatus is immersed in water, the wavelength of light changes according to the refractive index of water: $$\lambda' = \frac{\lambda}{\mu}$| \nConsequently, the new fringe width ($\beta'$) becomes: $$\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$| \nSubstituting the values: $$\beta' = \frac{1.178\text{ mm}}{4/3} = \frac{1.178 \times 3}{4}$| $$\beta' = \frac{3.534}{4} = 0.8835\text{ mm}$| \nTherefore, the new fringe width in water is $0.8835\text{ mm}$, and the reduction in fringe width is: $$\Delta \beta = \beta - \beta' = 1.178\text{ mm} - 0.8835\text{ mm} = 0.2945\text{ mm}$$

Question

In Young's double-slit experiment, monochromatic light of wavelength $589	ext{ nm}$ is used. The distance between the two slits is $0.5	ext{ mm}$ and the screen is kept at a distance of $1.0	ext{ m}$ from the slits. Calculate: The fringe width (distance between two consecutive bright fringes). The angular fringe width in radians. What will be the change in fringe width if the entire apparatus is immersed in water of refractive index $4/3$?

Accepted Answer

Given Data: Wavelength of light in air ($\lambda$) = $589	ext{ nm} = 589 	imes 10^{-9}	ext{ m}$ Distance between slits ($d$) = $0.5	ext{ mm} = 0.5 	imes 10^{-3}	ext{ m} = 5 	imes 10^{-4}	ext{ m}$ Distance of screen from slits ($D$) = $1.0	ext{ m}$ Refractive index of water ($\mu$) = $4/3 = 1.33$ --- Part 1: Calculation of Fringe Width ($\beta$)
The formula for fringe width in Young's double-slit experiment is: $$\beta = \frac{\lambda D}{d}$$ 
Substituting the given values: $$\beta = \frac{(589 	imes 10^{-9}	ext{ m}) 	imes (1.0	ext{ m})}{5 	imes 10^{-4}	ext{ m}}$$ $$\beta = \frac{589 	imes 10^{-9}}{5 	imes 10^{-4}}$$ $$\beta = 117.8 	imes 10^{-5}	ext{ m} = 1.178 	imes 10^{-3}	ext{ m} = 1.178	ext{ mm}$| --- Part 2: Calculation of Angular Fringe Width ($	heta$)
The formula for angular fringe width is: $$	heta = \frac{\beta}{D} = \frac{\lambda}{d}$| 
Substituting the values: $$	heta = \frac{589 	imes 10^{-9}	ext{ m}}{5 	imes 10^{-4}	ext{ m}}$$ $$	heta = 117.8 	imes 10^{-5}	ext{ radians} = 1.178 	imes 10^{-3}	ext{ rad}$$ --- Part 3: Change in Fringe Width in Water
When the apparatus is immersed in water, the wavelength of light changes according to the refractive index of water: $$\lambda' = \frac{\lambda}{\mu}$| 
Consequently, the new fringe width ($\beta'$) becomes: $$\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$| 
Substituting the values: $$\beta' = \frac{1.178	ext{ mm}}{4/3} = \frac{1.178 	imes 3}{4}$| $$\beta' = \frac{3.534}{4} = 0.8835	ext{ mm}$| 
Therefore, the new fringe width in water is $0.8835	ext{ mm}$, and the reduction in fringe width is: $$\Delta \beta = \beta - \beta' = 1.178	ext{ mm} - 0.8835	ext{ mm} = 0.2945	ext{ mm}$$