Question : A beam of light consisting of two wavelengths, 6500 Å and 5200 Å is used to obtain slit experiment. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.
(i) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å
(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide?
Doubt by Tushar
Solution :
λ1=6500 Å
λ2=5200 Å
d = 2 mm = 2×10-3 m
D = 120 cm =1.2 m
i) λ1=6500 ×10-10 m
We know,
xn=nλD/d
For third bright fringe, n=3
x3=3λD/d
x3=[3(6500 ×10-10)(1.2)]/(2×10-3)
x3 = [23400×10-10]/[2×10-3]
x3= 11700×10-7
x3 = 0.117 ×10-2 m
xn=nλD/d
For third bright fringe, n=3
x3=3λD/d
x3=[3(6500 ×10-10)(1.2)]/(2×10-3)
x3 = [23400×10-10]/[2×10-3]
x3= 11700×10-7
x3 = 0.117 ×10-2 m
ii) For least distance, we have to assume that the nth bright fringe due to light of wavelength λ1 coincide with the (n+1)th bright fringe due to the light of wavelength λ2.
Hence,
x=nλ1D/d=(n+1)λ2D/d
nλ1D/d=(n+1)λ2D/d
nλ1=(n+1)λ2
n/(n+1)=λ2/λ1
n/(n+1)=5200/6500
n/(n+1)=52/65
n/(n+1)=4/5
5n=4n+4
5n-4n=4
n=4
Hence, the 4th bright fringe due to wavelength λ1 get coincides with the 5th bright fringe due to wavelength λ2.
x=nλ1D/d=(n+1)λ2D/d
nλ1D/d=(n+1)λ2D/d
nλ1=(n+1)λ2
n/(n+1)=λ2/λ1
n/(n+1)=5200/6500
n/(n+1)=52/65
n/(n+1)=4/5
5n=4n+4
5n-4n=4
n=4
Hence, the 4th bright fringe due to wavelength λ1 get coincides with the 5th bright fringe due to wavelength λ2.
Required least distance will be
x=nλ1D/d
x = [4(6500×10-10)(1.2)]/[2×10-3]
x = [4.8×65×10-8]/[2×10-3]
x = 156×10-8+3
x = 156×10-5 m
x = [4(6500×10-10)(1.2)]/[2×10-3]
x = [4.8×65×10-8]/[2×10-3]
x = 156×10-8+3
x = 156×10-5 m
