Question 289: Wave Optics

Question

In Young's double slit experiment, the slits are separated by 0.6 mm and screen is placed at a distance of 1.2 m from slit. It is observed that the tenth bright fringe is at a distance of 8.85 mm from the third dark fringe on the same side. The wavelength of light used is

Accepted Answer

In Young's double slit experiment, the separation between the slits is 0.6 mm, and the screen is positioned 1.2 m away from the slits. The observed distance from the third dark fringe to the tenth bright fringe on the same side is 8.85 mm. We need to determine the wavelength of the light used.
Explanation
Formula for nth Bright Fringe:
$ y_n = \frac{n \lambda D}{d} $
For the 10th bright fringe:
$ y_{10} = \frac{10 	imes \lambda 	imes 1.2}{0.6 	imes 10^{-3}} = (20 	imes 10^3) \lambda \quad	ext{(Equation i)} $
Formula for nth Dark Fringe:
$ y_n' = \frac{(2n-1) \lambda D}{2d} $
For the 3rd dark fringe:
$ y_3' = \frac{5 \lambda 	imes 1.2}{2 	imes (0.6 	imes 10^{-3})} = (5 	imes 10^3) \lambda \quad	ext{(Equation ii)} $
Given:
$ y_{10} - y_3' = 8.85 	imes 10^{-3} $
Substituting from Equations (i) and (ii):
$ (20 	imes 10^3) \lambda - (5 	imes 10^3) \lambda = 8.85 	imes 10^{-3} $
Simplifying:
$ 15 	imes 10^3 \lambda = 8.85 	imes 10^{-3} $
Solving for $\lambda$:
$ \lambda = \frac{8.85 	imes 10^{-3}}{15 	imes 10^3} = 5.9 	imes 10^{-7} 	ext{ m} $
When expressed in Angstroms:
$ \lambda = 5900 \, 	ext{ÃÂ} $

Answer

5440 $$\mathop A\limits^o $$

Answer

5890 $$\mathop A\limits^o $$

Answer

5900 $$\mathop A\limits^o $$

Answer

6630 $$\mathop A\limits^o $$

Wave Optics

In Young's double slit experiment, the slits are separated by 0.6 mm and screen is placed at a distance of 1.2 m from slit. It is observed that the tenth bright fringe is at a distance of 8.85 mm from the third dark fringe on the same side. The wavelength of light used is

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