Question:
In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Answer:
Distance between the slits, d = 0.28 mm = 0.28 x 10-3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe,
u = 1.2 cm = 1.2 x 10-3 m
In case of a constructive interference, the relation for the distance between the two fringes is as
u = nλD/d Where, Order of fringes (n) = 4, Wavelength of light=λ
Therefore,
λ=ud/nD
= 1.2 x 10-3 x 0.28 x 10-3/ 4 x 4.14
= 6 x 10-7
= 600 nm
Hence, the wavelength of the light is 600 nm.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
