Science > Physics > Interference of Light > Numerical Problems on Bright and Dark Bands
In this article we shall study problems, to determine the nature of interference at a point i.e. to decide whether the point is a bright point or dark point.
Example – 01:
The optical path difference between two identical waves arriving at a point is 371λ. Is the point bright or dark? If the path difference is 0.24 mm, calculate the wavelength of light used.
Given: Path difference = 371λ, path difference = 0.24 mm = 0.24 x 10-3 m = 2.4 x 10-4 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = 371λ = 742 (λ/2)
Thus the path difference is even multiple of (λ/2)
Thus constructive interference takes place at the point. Hence the point is the bright point.
Given 371λ = 2.4 x 10-4
∴ λ = (2.4/371) x 10-4 = 6.469 x 10-7 m
∴ λ = 6469 x 10-10 m = 6469 Å
Ans: The point is a bright point and wavelength of light used is 6469 Å
Example – 02:
The optical path difference between two identical waves arriving at a point is 93 wavelengths. Is the point bright or dark? If the path difference is 46.5 micron, calculate the wavelength of light used.
Given: Path difference = 371λ, path difference = 0.24 mm = 46.5 micron = 46.5 x 10-6 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = 93λ = 186 (λ/2)
Thus the path difference is even multiple of (λ/2)
Thus constructive interference takes place at the point. Hence the point is the bright point.
Given 93λ = 46.5 x 10-6
∴ λ = (46.5/93) x 10-6 = 5 x 10-7 m
∴ λ = 5000 x 10-10 m = 5000 Å
Ans: The point is a bright point and wavelength of light used is 5000 Å
Example – 03:
The optical path difference between two identical waves arriving at a point is 85.5λ. Is the point bright or dark? If the path difference is 42.5 µm, calculate the wavelength of light used.
Given: Path difference = 85.5λ, path difference = 42.5 µm = 42.5 x 10-6 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = 85.5λ = 171 (λ/2)
Thus the path difference is the odd multiple of (λ/2)
Thus destructive interference takes place at the point. Hence the point is a dark point.
Given 85.5 λ = 42.5 x 10-6
∴ λ = (42.5/85.5) x 10-6 = 4.970 x 10-7 m
∴ λ = 4970 x 10-10 m = 4970 Å
Ans: The point is a dark point and wavelength of light used is 4970 Å
Example – 04:
The optical path difference between two identical waves arriving at a point is 6 x 10-6 m. The wavelength of light used is 5000 Å. Is the point bright or dark? What is the number of the band?
Given: Path difference = 6 x 10-6 m, Wavelength of light used = 5000 Å = 5000 x 10-10 m = 5 x 10-7 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = kλ = 6 x 10-6
∴ k x 5 x 10-7 = 6 x 10-6
∴ k = (6 x 10-6) / (5 x 10-7)= 12
∴ Path difference = 12λ = 24(λ /2)
Thus the path difference is even multiple of (λ/2)
Thus constructive interference takes place at the point. Hence the point is 12th bright point.
Ans: The point is 12th bright point
Example – 05:
The optical path difference between two identical waves arriving at a point is 185.5λ. Is the point bright or dark? If the path difference is 8.348 x 10-3 cm, calculate the wavelength of light used.
Given: Path difference = 85.5λ, path difference = 8.348 x 10-3 cm = 8.348 x 10-5 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = 185.5λ = 371 (λ/2)
Thus the path difference is an odd multiple of (λ/2)
Thus destructive interference takes place at the point
Hence the point is a dark point.
Given 185.5 λ = 8.348 x 10-5
∴ λ = (8.348/185.5) x 10-5 = 4.5 x 10-7 m
∴ λ = 4500 x 10-10 m = 4500 Å
Ans: The point is a dark point and wavelength of light used is 4500 Å
Example – 06:
The optical path difference between two identical waves arriving at a point is 13λ. Is the point bright or dark? If the path difference is 0.0078 mm, calculate the wavelength of light used.
Given: Path difference = 13λ, path difference = 0.0078 mm = 0.0078 x 10-3 m = 7.8 x 10-6 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = 13λ = 26 (λ/2)
Thus the path difference is even multiple of (λ/2)
Thus constructive interference takes place at the point. Hence the point is a bright point.
Given 13λ = 7.8 x 10-6
∴ λ = (7.8/13) x 10-4 = 6 x 10-7 m
∴ λ = 6000 x 10-10 m = 6000 Å
Ans: The point is a bright point and wavelength of light used is 6000 Å
Example – 07:
The optical path difference between two identical waves arriving at a point is 165.5λ. Is the point bright or dark? If the path difference is 9.75 x 10-5 m, calculate the wavelength of light used.
Given: Path difference = 371λ, path difference = 9.75 x 10-3 m
To Find: Nature of illumination and wavelength λ =?
Solution:
Path difference = 165.5λ = 331 (λ/2)
Thus the path difference is an odd multiple of (λ/2)
Thus destructive interference takes place at the point. Hence the point is a dark point.
Given 165.5λ = 9.78 x 10-5
∴ λ = (9.78 /165.5) x 10-5 = 5.891 x 10-7 m
∴ λ = 5891 x 10-10 m = 5891 Å
Ans: The point is a dark point and wavelength of light used is 5891 Å
Example – 08:
The slits in the interference experiment are illuminated by the light of wavelength 5600 Å. Find the path difference and a phase difference of the waves arriving at a point P on the screen to form 8th dark fringe.
Given: Wavelength of light used = 5600 Å = 5600 x 10-10 m = 5.6 x 10-7 m, 8th dark band, n = 8
To Find: Path difference and phase difference =?
Solution:
For dark band the path difference = (2n – 1) λ/2
∴ Path difference = (2 x 8 – 1)λ/2 = 7.5 λ = 7.5 x 5.6 x 10-7 = 4.2 x 10-6 m
Thus the phase difference = (2n – 1) π = (2 x 8 – 1)π = 15π
Ans: The path difference is 4.2 x 10-6 m and phase difference is 15π
Example – 09:
A point is situated at 6.5 cm and 6.65 cm from two coherent sources. Find the nature of the illumination of the point if wavelength of light is 5000 Å
Given: Distance of points from sources = 6.5 cm and 6.65 cm, wavelength of light used = λ = 5000 Å = 5000 x 10-10 m = 5 x 10-7 m
To Find: Nature of illumination =?
Solution:
Path difference = 6.65 cm – 6.5 cm = 0.15 cm = 0.15 x 10-2 m = 1.5 x 10-3 m
∴ Path difference = kλ = 1.5 x 10-3
∴ k x 5 x 10-7 = 1.5 x 10-3
∴ k = (1.5 x 10-3) / (5 x 10-7)= 3000
∴ Path difference = 3000λ = 6000(λ /2)
Thus the path difference is even multiple of (λ/2), Thus constructive interference takes place at the point. Hence the point is a bright point.
Ans: The point is a bright point
Example – 10:
A point P is situated at 8.7 cm and 8.8 cm from two coherent light sources. Is the point bright or dark? l = 5000 A.U.
Given: Distance of points from sources = 8.7 cm and 8.8 cm, wavelength of light used = λ = 5000 Å = 5000 x 10-10 m = 5 x 10-7 m
To Find: Nature of illumination =?
Solution:
Path difference = 8.8 cm – 8.7 cm = 0.1 cm = 0.1 x 10-2 m = 1 x 10-3 m
Path difference = kλ = 1 x 10-3
∴ k x 5 x 10-7 = 1 x 10-3
∴ k = (1.5 x 10-3) / ( 5 x 10-7)= 2000
∴ Path difference = 2000λ = 4000(λ /2)
Thus the path difference is even multiple of (λ/2), Thus constructive interference takes place at the point. Hence the point is a bright point.
Ans: The point is a bright point
Example – 11:
A screen is placed at a distance from two images of an illuminated slit. Distances of a point on the screen from the two images are 1.8 x 10-5 m and 1.23 x 10-5 m. If the wavelength of light used is 6000 A.U., is the point bright or dark? What is the number of bright or dark fringe formed at the point?
Given: Distance of points from sources = 1.8 x 10-5 m and 1.23 x 10-5 m, wavelength of light used = λ = 6000 Å = 6000 x 10-10 m = 6 x 10-7 m
To Find: Nature of illumination =?
Solution:
Path difference = 1.8 x 10-5 – 1.23 x 10-5 m = 0.57 x 10-5 m = 5.7 x 10-6 m
Path difference = kλ = 5.7 x 10-6
∴ k x 6 x 10-7 = 5.7 x 10-6
∴ K = (5.7 x 10-6)/( 6 x 10-7)= 9.5
∴ Path difference = 9.5λ = 19(λ /2)
∴ 2n – 1 = 19
∴ 2n = 20
∴ n = 10
Thus the path difference is even multiple of (λ/2), Thus constructive interference takes place at the point. Hence the point is a 10th dark point.
Ans: The point is a 10th bright point
Example – 12:
In Young’s experiment, the wavelength of monochromatic light used is 6000 Å. The optical path difference between the rays from the two coherent sources at point P on the screen is 0.0075 mm and at point Q on the screen is 0.0015 mm. How many bright and dark bands are observed between the two points P and Q. (Points P and Q are on the opposite sides of the central bright band)
Given: Wavelength of light used = λ = 6000 Å = 6000 x 10-10 m = 6 x 10-7 m
To Find: Number of bright and dark bands between P and Q.
Solution:
Consider point P:
Path difference = 0.0075 mm = 0.0075 x 10-3 m = 7.5 x 10-6 m
Path difference = kλ = 7.5 x 10-6
∴ k x 6 x 10-7 = 7.5 x 10-6
∴ k = (7.5 x 10-6) / (6 x 10-7)= 12.5
∴ Path difference = 12.5λ = 25(λ /2)
∴ (2n – 1) = 25
∴ 2n = 26
∴ n = 13
Thus the path difference is odd multiple of (λ/2). Thus destructive interference takes place at the point. Hence the point P is 13th dark point.
Consider point Q:
Path difference = 0.0015 mm = 0.0015 x 10-3 m = 1.5 x 10-6 m
Path difference = kλ = 1.5 x 10-6
∴ k x 6 x 10-7 = 1.5 x 10-6
∴ k = (1.5 x 10-6) / (6 x 10-7)= 2.5
∴ Path difference = 2.5λ = 5(λ /2)
∴ (2n – 1) = 5
∴ 2n = 6
∴ n = 3
Thus the path difference is an odd multiple of (λ/2). Thus destructive interference takes place at the point. Hence the point P is 3rd dark point.
Points P and Q are on the opposite sides of the central bright band
Hence the number of dark bands between P and Q
= 13 + 3 = 16 (including those at P and Q) Or 14 (excluding those at P and Q)
Hence the number of bright bands between P and Q
= (13 -1) + (3 – 1) + 1 central band = 12 + 2 + 1 = 15
Ans: Number of bright bands = 15 and number of dark bands = 14
Example – 13:
In Young’s experiment, the wavelength of monochromatic light used is 4000 Å. The optical path difference between the rays from the two coherent sources at point P on the screen is 0.0080 mm and at point Q on the screen is 0.0022 mm. How many bright and dark bands are observed between the two points P and Q. (Points P and Q are on the opposite sides of the central bright band)
Given: Wavelength of light used = λ = 4000 Å = 4000 x 10-10 m = 6 x 10-7 m
To Find: Number of bright and dark bands between P and Q.
Solution:
Consider point P:
Path difference = 0.0080 mm = 0.0080 x 10-3 m = 8 x 10-6 m
Path difference = kλ = 8 x 10-6
∴ k x 4 x 10-7 = 8 x 10-6
∴ k = (8 x 10-6) / (4 x 10-7) = 20
Path difference = 20λ = 40(λ /2)
n = 20
Thus the path difference is even multiple of (λ/2). Thus constructive interference takes place at the point. Hence the point P is 20th bright point.
Consider point Q:
Path difference = 0.0022 mm = 0.0022 x 10-3 m = 2.2 x 10-6 m
Path difference = kλ = 2.2 x 10-6
∴ k x 4 x 10-7 = 2.2 x 10-6
∴ k = (2.2 x 10-6) / (4 x 10-7)= 5.5
∴ Path difference = 5.5λ = 11(λ /2)
∴ (2n – 1) = 11
∴ 2n = 12
∴ n = 6
Thus the path difference is an odd multiple of (λ/2). Thus destructive interference takes place at the point. Hence the point P is 6th dark point.
Points P and Q are on the opposite sides of the central bright band
Hence the number of bright bands between P and Q
= 20 + 1 central band + (6 – 1) = 21 + 5 = 26 (including that at P) Or 25 (excluding that at P)
The number of dark bands between P and Q
= 20 + 6 = 26 (including that at Q) 0r 25 (excluding that at Q)
Ans: Number of bright bands = 25 and number of dark bands = 25
Example – 14:
A phase difference between the coherent waves of wavelength λ originating from two slits is 3π radian. A point P on the screen is at a distance of 20λ and 21.5λ from the two slits. Is point P bright or dark?
To Find: Nature of illumination =?
Solution:
Path difference = 21.5λ – 20λ = 1.5λ
Phase difference of source = 3π radian = (3/2)(2π) radian
2π radian corresponds to λ
Hence 3π radian corresponds to 1.5λ
Thus the phase difference of waves meeting at P is 1.5λ ± 1.5λ = 3λ or 0
In either case, it is an integral multiple of λ. Hence point P is the bright point.
Ans: Point P is a bright point
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