|Class 12 Physics Wave Optics||Diffraction|
Diffraction Fringe Pattern
Diffraction pattern Maxima
Problem:-The light of wavelength 600nm is incident normally on a slit of width 3mm.Calcluate the linear width of central maximum on a screen kept 3m away from the slit?
Wavelength λ =600nm =600 x10-9m.
Width of the slit a =3mm=3x10-3m.
Distance of the screen D=3m.
Condition for minima:-a sinθ =nλ
=>a sin θ =λ
=>sin θ = (λ/a) = (600 x10-9)/ (3x10-3)
=>θ =(600 x10-9)/(3x10-3)
=>(x/D) = (600 x10-9)/ (3x10-3)
=> x= (600 x 10-9 x3)/ (3 x10-3)
x=600 x 10-6m.
Therefore width =2x = 1200 x 10-6m
Width = 1.2 mm
Diffraction pattern Secondary Maxima
Scientist Fresnel found that secondary maxima occurred when the value of θ is:
Slit divided into 3 parts
Diffraction pattern for Minima
Problem:- A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtaininterference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the centralmaximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes dueto both the wavelengths coincide?
Wavelength of the light beam, λ1 =650nm
Wavelength of another light beam,λ2 =520nm
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 bythe relation,
For third bright fringe, n=3
Therefore x=3x650(D/d) =1950(D/d) nm
(b) Let the nth bright fringe due to wavelength and (n − 1)th bright fringe due towavelength λ1coincide on the screen. We can equate the conditions for bright fringesas:
Hence, the least distance from the central maximum can be obtained by the relation:
Note: The value of d and D are not given in the question.
Problem:- In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screenplaced 1 m away. The wavelength of light used is 600 nm. What will be the angularwidthof the fringe if the entire experimental apparatus is immersed in water? Take refractiveindex of water to be 4/3.
Distance of the screen from the slits, D = 1 m
Wavelength of light used, λ1 =600nm
Angular width of the fringe in air θ1 =0.2o
Angular width of the fringe in water = θ2
Refractive index of water, μ= (4/3)
Refractive index is related to angular width as:
μ = (θ1/θ2)
θ2 = (3/4)θ1
= (3/4)x0.2 =0.15
Therefore, the angular width of the fringe in water will reduce to 0.15°.
Problem:- A parallel beam of light of wavelength 500 nm falls on a narrow slitand the resulting diffraction pattern is observed on a screen 1 maway. It is observed that the first minimum is at a distance of 2.5mm from the centre of the screen. Find the width of the slit.
Wavelength of light beam, λ = 500 nm = 500 × 10−9 m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 × 10−3 m
It is related to the order of minima as:
n λ =x(d/D)
d= (n λD/x)
d= (1x500x10-9x1)/ (2.5 x10-3) = 2x 10-4 m
Therefore, the width of the slits is 0.2 mm.