In Young’s double-slit experiment,


In Young’s double-slit experiment, 0.28mm separation between the slits and the screen is placed 1.4m away. 1.2cm is the distance between the central bright fringe and the fourth bright fringe. Determine the wavelength of light used in the experiment.


Distance between the slits and the screen, D = 1.4 m

and the distance between the slits, d = 0.28 mm = 0.28 x 10-3 m

Distance between the central fringe and the fourth (n = 4) fringe,

u = 1.2cm = 1.2 × 10-2 m

For constructive interference, the following is the relation for the distance between the two fringes:

$u=n \lambda \frac{D}{d}$

Where, n = order of fringes

$=4 \lambda=$ Wavelength of light used

Rearranging the formula, we get

$\lambda=\frac{u d}{n D}$

$=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$

= 6 × 10-7 m = 600nm


600nm is the wavelength of the light.



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