This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Conditions required for interference of light:1. Light waves from coherent sources2. Same or almost the same amplitudes3. Small separation between two coherent sources4. Small path difference
Move slit S and light source to double slit- bright fringes are brighter (intensity of light through double slit is greater)- fringe separation x unchanged
From A to O, optical path length = AOFrom B to O, optical path length = BO + ( - 1)d = AO + ( - 1)d Optical path difference = [AO - ( - 1)d] – AO = ( - 1)d 0
Central maximum shifts to O’
If ( - 1)d = n (the nth bright fringe at O’)If ( - 1)d = (n - ) (the nth dark fringe at O’)
Light are plane polarized- No interference pattern (lights are mutually perpendicular)- or visible interference pattern (planes of polarization are parallel)
Q: Q: (a) What is meant by interference?(b) State the conditions necessary for the production of an interference pattern.(c) State the conditions necessary to produce an interference pattern where the fringes are of the same separation and of about the same intensity in a Young’s double slit experiment.(d) In a Young’s double slit experiment, light of wavelength 644 nm was used. Initially, the slit separation is 0.20 mm and the distance between the screen and the double slit is 1.00 m. Throughout the experiment, the interference pattern was viewed from the position of the double slit.(i) What is the angular separation between two neighbouring bright fringes formed on the screen?(ii) If the screen is then shifted further away from the double slit, what changes, if any, will occur to the interference pattern?
(a) Interference is the effect produced by the superposition of waves from two coherent sources moving through the same region.
(b) The conditions necessary for the production of an interference pattern are:1. The waves must be from two coherent sources.2. The amplitudes of the waves must be the same or almost the same.3. The two coherent sources must be close to each other.
The conditions necessary to produce evenly spaced fringes are:1. The use of a monochromatic source.2. The single slit S must be narrow and lie on the perpendicular divider of AB (see figure above).3. The slit separation a must be small.4. The distance of the screen from the double slit, D, must be large.
Q: Q: (a) What is meant by Huygens’ principle?(b) With the aid of a diagram, use Huygens’ principle to explain how the interference pattern is produced in a Young’s double slit experiment.(c) In a Young’s double slit experiment, the slit separation is 0.05 cm and the distance between the double slit andscreen is 200 cm. When blue light is used, the distance of the first bright fringe from the centre of the interferencepattern is 0.13 cm.(i) Calculate the wavelength of the blue light used in the experiment.(ii) Calculate the distance of the fourth dark fringe from the centre of the interference pattern.
Spherical wavefronts emerge from the slit S. Since slits A and B are of an equal distance from S, the same wavefront from S arrives at A and B simultaneously. According to Huygens’ Principle, the points on the wavefront at A and B emit wavelets in phase with one other. Hence, the waves from A and B are coherent.
At the points where the optical path difference of waves from A and B is nλ ( n = 0, 1, 2...), constructive interference occurs and a bright fringe is obtained. If the optical path difference is ( n – )λ, then destructive interference occurs and a dark fringe is obtained.
Q:Q: A Young’s double slit system was set up at one end of an empty rectangular glass tank, and a sodium lamp with a narrow slit placed outside the tank. Interference fringes were observed on the opposite side of the tank. When the tank was filled with oil, the fringe separation changes by 35%. Explain why this happens and discuss whether the fringe separation increases or decreases. Calculate the refractive index of the oil. Solution
Q:Q: In a Young’s double slit experiment, an interference pattern was obtained using monochromatic light of wavelength 500 nm. When a thin film of transparent material with refractive index 1.5 was placed in front of one of the slits, the central maximum was shifted to the original position of the 9th bright fringe. What is the thickness of the film?
When the central maximum was shifted to O’, the original position of the 9th bright fringe above O, all the other fringes were also shifted upwards. Therefore, at the point O on the axis of the double slit system, we will find the 9th bright fringe which was originally below O.
Q:Q: In a Young’s double slit arrangement, the distance
between the centres of the slits is 0.25 mm and light of wavelength 6.0 × 10–4 mm is used. Calculate the angle θ subtended at the double slit by the neighbouring maxima of the interference pattern (refer to figure below).
interference fringes if(a) the slit A is covered with a thin piece of transparent material of high refractive index,(b) the intensity of light from A is reduced to half of that from B,(c) A and B are covered with thin films of polaroid, and one of the films is rotated slowly,(d) the distance between A and B is increased slowly.
Solution (cont’d):Solution (cont’d):(a) When the slit A is covered with a thin piece of transparent material of high refractive index, the central maximum which was originally at O is shifted upwards.When A is covered with the transparent material, the optical path for light travelling from A to O is AO + (μ – 1)dwhere μ is the refractive index of the material and d its thickness.Hence the optical path difference between light travelling from A to O and from B to O is optical path difference = [AO + (μ – 1)d] – OB = (μ – 1) d (AO = OB) ≠ 0Hence O is no more the position of the central maximum.If (μ – 1)d = nλ where n is an integerthen the nth bright fringe from central maximum will be formed at O.If (μ – 1)d = (n – )λthen the nth dark fringe from the central maximum is found at O.
Solution (cont’d):Solution (cont’d):(b) Initially, the intensity of the light across the interference pattern is as shown in Fig. (a). When the intensity of the light from A is halved, the intensity of the light across the interference pattern is as shown in Fig. (b).
Fig. (a)
Fig. (b)The intensity of the maxima has decreased but the intensity of the minima has increased. Hence the difference between the intensities of the maxima and minima is reduced. There is no change in the fringe separation.
(c) When the axis of polarization of the polaroids are parallel, the bright fringes can be differentiated clearly for the dark fringes. When one of the polaroids is rotated, the difference in intensities of the maxima and minima decreases until the axes of the polaroids are mutually perpendicular. Then no interference pattern will be seen.
(d) Using the equation x = ,
when the slit separation a is slowly increased the slit separation x decreases. The fringes become closer and closer until the neighbouring bright fringes cannot be resolved by the eye.
Q:Q: In a Young’s double slit experiment, a source of light which emits light of wavelengths 7 × 10–7 m (red) and 5.6 × 10–7 m (green) is used. Draw a sketch to show the interference fringes of the two colours formed. Show that the fringes ofthe two colours first exactly overlap while moving out from the centre of the pattern when
What is the significance of n? What is the colour of the fringe at that point?
Solution:Solution: For red light, the fringe separation
= (7 × 10–7)
For green light, the fringe separation = (5.6 × 10–7) x1
When a red fringe and a green fringe overlap for the first time, the nth red fringe overlap with the (n + 1)th green fringe; and their distances from the central maximum are equal.i.e. nx1 = (n + 1) x2
Solution (cont’d):Solution (cont’d):n represents the fourth red fringe which coincides with the fifth green fringe. The colour of the fringe produced is
Note that for n = 0 and n = 4, a red and green fringe coincide exactly to produce a yellow fringe.