Question

In Young's double slits experiment, the position of $5^{\text {th }}$ 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 $600 \,mm$ The separation between the slits is:

• A. $12\, \mu m$
• B. $60\, \mu m$
• C. $36\, \mu m$
• D. $48 \,\mu m$

Answer 1

The Correct Option is D

1. The position of bright fringes in Young's double-slit experiment is given by: \[ y_n = \frac{n \lambda D}{d}, \] where \(n\) is the fringe order, \(\lambda\) is the wavelength of light, \(D\) is the distance between the slits and screen, and \(d\) is the distance between the slits. 
2. Substituting the given values: - \(n = 5, \, \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}, \, y_n = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m}, \, D = 1 \, \text{m}\). 
3. Rearrange for \(d\): \[ d = \frac{n \lambda D}{y_n}. \] 
4. Substituting: \[ d = \frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}} = 6 \times 10^{-6} \, \text{m} = 48 \, \mu\text{m}. \] 
Thus, the distance between the slits is 48 \(\mu\text{m}\). 
The fringe spacing depends on the wavelength, the slit-to-screen distance, and the slit separation. Accurate calculations require converting all quantities to SI units.