A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s 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?
a)
Given: Wavelength of first light beam is 650 mm and wavelength of second light beam is 520 mm .
The distance of n th bright fringe on the screen from the central maximum is given as,
x n = n λ D d (1)
Where, D is the distance of slit from the screen, d is the distance between the slits, n is the number of bright fringe from the central maximum and λ is the wavelength of the light beam.
By substituting the value of n and λ in the above equation, we get
x 3 = 3 × 650 D d = 1950 ( D d ) nm
Here, value of D and d is not given.
Thus, the distance of third bright fringe on the screen from the central maximum is 1950 ( D d ) nm
b)
Let the n th bright fringe due to wavelength λ 1 and ( n − 1 ) th bright fringe due to wavelength λ 1 coincide on the screen. Then,
n λ 2 = ( n − 1 ) λ 1 n × 520 = ( n − 1 ) × 650 130 n = 650 n = 5
By substituting the value of n in the above equation (1), we get
x 5 = 5 × 520 D d = 2600 ( D d ) nm
Thus, the distance from the central bright maxima where the bright fringe for both the wavelengths coincide is 2600 ( D d ) nm .
a)
Given: Wavelength of first light beam is
The distance of
Where,
By substituting the value of
Here, value of
Thus, the distance of third bright fringe on the screen from the central maximum is
b)
Let the
By substituting the value of
Thus, the distance from the central bright maxima where the bright fringe for both the wavelengths coincide is
