Problems on Young’s Double Slit Experiment (YDSE) – Solved Numericals

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1. In Young’s experiment, the distance between the two slits is 0.8 mm and the distance of the screen from the slits is 1.2m. If the fringe width is 0.75 mm, calculate the wavelength of light.

Solution:

• Distance between the two slits (\(d\)) = 0.8 mm = \(0.8 \times 10^{-3} \, \text{m} = 8 \times 10^{-4} \, \text{m}\)
• Distance from the slits to the screen (\(D\)) = 1.2 m
• Fringe width (\(\beta\)) = 0.75 mm = \(0.75 \times 10^{-3} \, \text{m} = 7.5 \times 10^{-4} \, \text{m}\)

In Young’s double-slit experiment, the fringe width is given by:

\[ \beta = \frac{\lambda D}{d} \]

Rearranging for the wavelength (\(\lambda\)):

\[ \lambda = \frac{\beta d}{D} \] \[ \lambda = \frac{(7.5 \times 10^{-4}) \times (8 \times 10^{-4})}{1.2} \] \[ \lambda = \frac{60 \times 10^{-8}}{1.2} = 5 \times 10^{-7} \text{m} \] \[ \lambda = 5000 Å \]

2. In Young’s experiment, the distance between the two images of the sources is 0.6 mm and the distance between the source and the screen is 1.5 m. Given that the overall separation between 20 fringes on the screen is 3 cm, calculate the wavelength of light used.

Solution:

• Distance between the two slits (\(d\)) = 0.6 mm = \(0.6 \times 10^{-3} \, \text{m} = 6 \times 10^{-4} \, \text{m}\)
• Distance from the slits to the screen (\(D\)) = 1.5 m
• Total separation between 20 fringes = 3 cm = \(3 \times 10^{-2} \, \text{m}\)
• Number of fringes = 20
• \(\beta = 3/20 cm = 0.0015 m \)

In Young’s experiment, the fringe width is given by:

\[ \beta = \frac{\lambda D}{d} \]

Rearranging for the wavelength \(\lambda\):

\[ \lambda = \frac{\beta d}{D} \] \[ \lambda = \frac{0.0015 \times (6 \times 10^{-4})}{1.5} \] \[ \lambda = \frac{0.009 \times 10^{-4}}{1.5} \] \[ \lambda = \frac{9 \times 10^{-7}}{1.5} \] \[ \lambda = 6 \times 10^{-7} \, \text{m} \] \[ \lambda = 6000 Å \]