Question:
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Youngs double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Answer:
nλ2 = (n-1) λ1 Where, λ1 = 650 nm
n= 5
Wavelength of another light beam (λ2) = 520 nm, Distance of the slits from the screen = D , Distance between the two slits = d
(a)
Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
x = n λ1(D/d), If third bright fringe. N=3
Then,
x = 3 x 650D/d = 1950(D/d) nm
(b)
Let the nth bright fringe due to wavelength λ2 and (n – 1)th bright fringe due to wavelength λ1 coincide on the screen. We can equate the conditions for bright fringes as:
nλ2 = (n-1) λ1
520n = 650n - 650
n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
x = λ2D/d
= 5 x 520 x D/d
= 260 D/d nm
Not what you are looking for? Go ahead and submit the question, we will get back to you.
