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LIGHT WAVES
EXERCISES Q-31 TO Q-41
31. In a Young's double slit experiment 𝜆 = 500 nm, d = 1.0 mm and D = 1.0 m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
ANSWER: If the intensity of each of the slits =I', the resultant intensity I =4I'*cos²(ẟ/2)
where ẟ is the phase difference.
→cos²(ẟ/2) = I/4I'
Diagram for Q-31
For the central maximum, the intensity at the bright fringe =4I', because of ẟ =2nπ. Suppose at a distance y from the central maximum the intensity is half of the maximum intensity. i.e. I = 2I'.
→cos²(ẟ/2) = 2I'/4I' =½
→cos(ẟ/2) = 1/√2
→ẟ/2 = π/4
→ẟ = π/2 = phase difference.
→2π*Δx/𝜆 = π/2
→Δx = 𝜆/4
→yd/D =𝜆/4
→y = 𝜆D/4d = 500x10⁻⁹ *1.0/(4*0.001)
→y = 1.25x10⁻⁴ m
32. The line width of a bright fringe is some times defined as the separation between the points on the two sides the central line where the intensity falls to half the maximum. Find the line width of a bright fringe in a Young's double slit experiment in terms of 𝜆, d and D where the symbols have their usual meaning.
ANSWER: If the intensity of each of the slits =I', the resultant intensity I =4I'*cos²(ẟ/2)
where ẟ is the phase difference.
→cos²(ẟ/2) = I/4I'
For the central maximum, the intensity at the bright fringe =4I', because of ẟ =2nπ. Suppose at a distance y from the central maximum the intensity is half of the maximum intensity. i.e. I = 2I'.
→cos²(ẟ/2) = 2I'/4I' =½
→cos(ẟ/2) = 1/√2
→ẟ/2 = π/4
→ẟ = π/2 = phase difference.
→2π*Δx/𝜆 = π/2
→Δx = 𝜆/4
→yd/D =𝜆/4
→y = 𝜆D/4d
But the line width is the separation between the points on either side of the central maximum where the intensity falls to half of the maximum. Hence the line width = 2y = 2𝜆D/4d = 𝜆D/2d
33. Consider the situation shown in figure (17-E6). The two slits S₁ and S₂ placed symmetrically around the central line are illuminated by monochromatic light of wavelength 𝜆. The separation between the slits is d. The light transmitted by the slits falls on a screen Σ₁ placed at a distance D from the slits. The slit S₃ is at the central line and the slit S₄ is at a distance z from S₃. Another screen Σ₂ is placed a further distance D away from Σ₁. Find the ratio of the maximum to minimum intensity observed on Σ₂ if z is equal to
(a) 𝜆D/2d
(b) 𝜆D/d
(c) 𝜆D/4d
Figure for Q-33
ANSWER: Let the intensity of each of the slits S₁ and S₂ =I'. The corresponding amplitude = r (say). S₃ is at the center of the screen where the path difference and hence the phase difference is zero. So here the intensity is maximum = I (say). Hence I = 4I'.
At S₄, the path difference Δx = zd/D
Phase difference ẟ = 2π*Δx/𝜆 =2πzd/𝜆D
(a) For z =𝜆D/2d
ẟ = (2πd/𝜆D)*(𝜆D/2d) =π
Since at S₄, the phase difference is π, the destructive interference takes place and the minima occur where the intensity is zero.
So for the screen Σ₂, out of the two slits S₃ and S₄, only S₃ is illuminated with an intensity I. The screen will be uniformly illuminated having the same value for maxima and minima. Hence their ratio = 1
(b) For z =𝜆D/d
ẟ = (2πd/𝜆D)*(𝜆D/d) =2π
Since at S₄, the phase difference is 2π, the constructive interference takes place and the maxima occur where the intensity is I same as S₃.
So for the screen Σ₂, the two slits S₃ and S₄ are illuminated with an intensity I. They will form an interference pattern with bright and dark lines i.e. with maximum intensity = I" (say) and minimum intensity zero. Hence the ratio = I"/0 = ∞.
(c) For z =𝜆D/4d
ẟ = (2πd/𝜆D)*(𝜆D/4d) =π/2
Since at S₄, the phase difference = π/2, the intensity here is given as
I₁ = 4I'.cos²(ẟ/2) =4I'.cos²(π/4) =4I'*½ =2I'.
So S₃ is illuminated with an intensity I = 4I' and S₄ with intensity 2I'.
Let the amplitude for I' = r, for 2I' = r₁ and for 4I' =r₂. So,
2I'/I' = r₁²/r²
→r₁²/r² =2
→r₁² = 2r²
→r₁ =√2.r
And, 4I'/I' = r₂²/r²
→r₂²/r² = 4
→r₂² = 4r²
→r₂ = 2r
Hence the amplitude of the light wave coming from S₃ is 2r and that coming from S₄ is √2.r, so when they interfere the amplitude of the maximum intensity (say Iₘ) =2r+√2.r and the amplitude of the minimum intensity (say Iₙ) = 2r-√2r. Thus the ratio of maximum to minimum intensity on the screen Σ₂
Iₘ/Iₙ = (2r+√2.r)²/(2r-√2.r)²
=(2+√2)²/(2-√2)²
=(4+2+4√2)/(4+2-4√2)
=(6+4√2)/(6-4√2)
=(3+2√2)/(3-2√2)
Multiply numerator and denominator by (3+2√2)
→Iₘ/Iₙ = (3+2√2)²/(3²-8) =(9+8+12√2)/(9-8)
=(17+12√2)/1 =17+12*1.41 =17+16.92 ≈34
34. Consider the arrangement shown in figure (17-E7). By some mechanism, the separation between the slits S₃ and S₄ can be changed. The intensity at the point P which is at the common perpendicular bisector of S₁S₂ and S₃S₄. When z = D𝜆/2d, the intensity measured at P is I. Find the intensity when z is equal to
(a) D𝜆/d
(b) 3D𝜆/2d, and
(c) 2D𝜆/d
Figure for Q-34
ANSWER: Let the intensity of S₁ and S₂ = I'. For S₃ and S₄, y = z/2. Hence the path difference here is
Δx = yd/D =zd/2D
=(D𝜆/2d)d/2D
=𝜆/4
The phase difference ẟ = 2π(Δx)/𝜆 =2π𝜆/4𝜆 =π/2.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π/4) =4I'*½ =2I'
where I' is the intensity of each of the sources. For the point P, the holes S₃ and S₄ are sources with intensity 2I' of each. Also, the point P is the point of central maxima, hence the intensity at P is
I = 4*2I' =8I'.
→I' = I/8.
(a) when z = D𝜆/d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (D𝜆/d)*d/2D =𝜆/2
The phase difference ẟ = 2π(Δx)/𝜆 =2π𝜆/2𝜆 =π.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π/2) =4I'*0 =0
So the intensities at each of the slits S₃ and S₄ is zero and will be dark. Therefore the intensity at the point P is also zero.
(b) when z = 3D𝜆/2d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (3D𝜆/2d)*d/2D =3𝜆/4
The phase difference ẟ = 2π(Δx)/𝜆 =2π(3𝜆/4)/𝜆 =6π/4 =3π/2.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(3π/4) =4I'*½ =2I'
For the point P, the slits S₃ and S₄ are sources with intensity 2I' each. Hence the intensity at the central point P where maxima forms is = 4*2I' = 8I' =I.
(c) when z = 2D𝜆/d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (2D𝜆/d)*d/2D =𝜆
The phase difference ẟ = 2π(Δx)/𝜆 =2π(𝜆)/𝜆 =2π
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π) =4I'*1 =4I'.
Thus the intensity at the maxima at P = 4*4I' =16I'
=16*(I/8) = 2I.
35. A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution, if it is known to be 1.2 and 1.5?
ANSWER: Given 𝜆 =580 nm =580x10⁻⁹ m
The thickness of the film, d =0.0011 mm = 1.1x10⁻⁶ m
Since the bubble appears dark when seen by the reflected light, destructive interference takes place, hence
2µd =n𝜆
→2µ =n*580x10⁻⁹/1.1x 10⁻⁶
→µ = n*0.264
Since it is known that µ is between 1.2 and 1.5, for the value of n = 5, µ = 5*0.264 =1.32.
36. A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index 1.4). What should be the minimum thickness of the film so that it strongly reflects the light?
ANSWER: Given 𝜆 =560 nm
µ = 1.4, Thickness d = ?
For the film to strongly reflect the light
2µd =(n+½)𝜆
→2*1.4*d = (n+½)*560
→d = (n+½)*200 nm
The minimum value of d will be for n = 0
→d = 200/2 =100 nm
37. A parallel beam of white light is incident normally on a water film 1.0x10⁻⁴ cm thick. Find the wavelength in the visible range (400 nm - 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33.
ANSWER: For the light to be strongly transmitted by the film,
2µd = n𝜆
→𝜆 = 2µd/n = 2*1.33*1.0x10⁻⁴*10⁹/100n
=2.66x1000/n =2660/n
For n = 4, 𝜆 = 665 nm
for n = 5, 𝜆 = 532 nm
for n = 6, 𝜆 = 443 nm
For other values of n, the 𝜆 is beyond the range 400 nm - 700 nm.
38. A glass surface is coated by an oil film of uniform thickness 1.0x10⁻⁴ cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelength of the light in the visible range (400 nm-750 nm) which are completely transmitted by the oil film under normal incidence.
ANSWER: For the oil thickness of the film, d = 1.0x10⁻⁴ cm, µ = 1.25.
For complete transmission under the normal incidence, 2µd = (n+½)𝜆.
→𝜆 = 2µd/(n+½)
=2*1.25*1.0x10⁻⁴*10⁷/(n+½)
=2.5x1000/(n+½) nm
=2500/(n+½) nm
For n = 3, 𝜆 = 2500/3.5 =714 nm,
n = 4, 𝜆 = 2500/4.5 =556 nm,
n = 5, 𝜆 = 2500/5.5 =455 nm.
For other values of n, 𝜆 is out of the given range.
39. Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30°.
ANSWER: The first diffraction minimum is given at
b.sinθ = 𝜆
→𝜆 = b.sinθ = 5.sin30° cm
=2.5 cm
40. Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and fall on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?
ANSWER: The bright spot is the circular area within the first dark ring. The radius of the central bright spot
R = 1.22𝜆D/b
Here 𝜆 = 560 nm =560x10⁻⁹ m
D = 2.0 m
b = 0.20 mm =0.0002 m
→R = 1.22*560x10⁻⁹*2/0.0002 m
=1.22*560x10⁻⁵*100 cm
≈ 0.68 cm
41. A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?
ANSWER: The distance D = 20 cm =0.20 m
𝜆 = 620 nm =620x10⁻⁹ m
The radius of the lens, b = 8/2 cm =4 cm =0.04 m
The radius of the central bright spot is given as
R = 1.22𝜆D/b
= 1.22*620x10⁻⁹*0.2/0.04 m
= 3782x10⁻⁹ m
≈ 3.8x10⁻⁶ m
31. In a Young's double slit experiment 𝜆 = 500 nm, d = 1.0 mm and D = 1.0 m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
ANSWER: If the intensity of each of the slits =I', the resultant intensity I =4I'*cos²(ẟ/2)
where ẟ is the phase difference.
→cos²(ẟ/2) = I/4I'
For the central maximum, the intensity at the bright fringe =4I', because of ẟ =2nπ. Suppose at a distance y from the central maximum the intensity is half of the maximum intensity. i.e. I = 2I'.
→cos²(ẟ/2) = 2I'/4I' =½
→cos(ẟ/2) = 1/√2
→ẟ/2 = π/4
→ẟ = π/2 = phase difference.
→2π*Δx/𝜆 = π/2
→Δx = 𝜆/4
→yd/D =𝜆/4
→y = 𝜆D/4d = 500x10⁻⁹ *1.0/(4*0.001)
→y = 1.25x10⁻⁴ m
32. The line width of a bright fringe is some times defined as the separation between the points on the two sides the central line where the intensity falls to half the maximum. Find the line width of a bright fringe in a Young's double slit experiment in terms of 𝜆, d and D where the symbols have their usual meaning.
ANSWER: If the intensity of each of the slits =I', the resultant intensity I =4I'*cos²(ẟ/2)
where ẟ is the phase difference.
→cos²(ẟ/2) = I/4I'
For the central maximum, the intensity at the bright fringe =4I', because of ẟ =2nπ. Suppose at a distance y from the central maximum the intensity is half of the maximum intensity. i.e. I = 2I'.
→cos²(ẟ/2) = 2I'/4I' =½
→cos(ẟ/2) = 1/√2
→ẟ/2 = π/4
→ẟ = π/2 = phase difference.
→2π*Δx/𝜆 = π/2
→Δx = 𝜆/4
→yd/D =𝜆/4
→y = 𝜆D/4d
But the line width is the separation between the points on either side of the central maximum where the intensity falls to half of the maximum. Hence the line width = 2y = 2𝜆D/4d = 𝜆D/2d
33. Consider the situation shown in figure (17-E6). The two slits S₁ and S₂ placed symmetrically around the central line are illuminated by monochromatic light of wavelength 𝜆. The separation between the slits is d. The light transmitted by the slits falls on a screen Σ₁ placed at a distance D from the slits. The slit S₃ is at the central line and the slit S₄ is at a distance z from S₃. Another screen Σ₂ is placed a further distance D away from Σ₁. Find the ratio of the maximum to minimum intensity observed on Σ₂ if z is equal to
(a) 𝜆D/2d
(b) 𝜆D/d
(c) 𝜆D/4d
ANSWER: Let the intensity of each of the slits S₁ and S₂ =I'. The corresponding amplitude = r (say). S₃ is at the center of the screen where the path difference and hence the phase difference is zero. So here the intensity is maximum = I (say). Hence I = 4I'.
At S₄, the path difference Δx = zd/D
Phase difference ẟ = 2π*Δx/𝜆 =2πzd/𝜆D
(a) For z =𝜆D/2d
ẟ = (2πd/𝜆D)*(𝜆D/2d) =π
Since at S₄, the phase difference is π, the destructive interference takes place and the minima occur where the intensity is zero.
So for the screen Σ₂, out of the two slits S₃ and S₄, only S₃ is illuminated with an intensity I. The screen will be uniformly illuminated having the same value for maxima and minima. Hence their ratio = 1
(b) For z =𝜆D/d
ẟ = (2πd/𝜆D)*(𝜆D/d) =2π
Since at S₄, the phase difference is 2π, the constructive interference takes place and the maxima occur where the intensity is I same as S₃.
So for the screen Σ₂, the two slits S₃ and S₄ are illuminated with an intensity I. They will form an interference pattern with bright and dark lines i.e. with maximum intensity = I" (say) and minimum intensity zero. Hence the ratio = I"/0 = ∞.
(c) For z =𝜆D/4d
ẟ = (2πd/𝜆D)*(𝜆D/4d) =π/2
Since at S₄, the phase difference = π/2, the intensity here is given as
I₁ = 4I'.cos²(ẟ/2) =4I'.cos²(π/4) =4I'*½ =2I'.
So S₃ is illuminated with an intensity I = 4I' and S₄ with intensity 2I'.
Let the amplitude for I' = r, for 2I' = r₁ and for 4I' =r₂. So,
2I'/I' = r₁²/r²
→r₁²/r² =2
→r₁² = 2r²
→r₁ =√2.r
And, 4I'/I' = r₂²/r²
→r₂²/r² = 4
→r₂² = 4r²
→r₂ = 2r
Hence the amplitude of the light wave coming from S₃ is 2r and that coming from S₄ is √2.r, so when they interfere the amplitude of the maximum intensity (say Iₘ) =2r+√2.r and the amplitude of the minimum intensity (say Iₙ) = 2r-√2r. Thus the ratio of maximum to minimum intensity on the screen Σ₂
Iₘ/Iₙ = (2r+√2.r)²/(2r-√2.r)²
=(2+√2)²/(2-√2)²
=(4+2+4√2)/(4+2-4√2)
=(6+4√2)/(6-4√2)
=(3+2√2)/(3-2√2)
Multiply numerator and denominator by (3+2√2)
→Iₘ/Iₙ = (3+2√2)²/(3²-8) =(9+8+12√2)/(9-8)
=(17+12√2)/1 =17+12*1.41 =17+16.92 ≈34
34. Consider the arrangement shown in figure (17-E7). By some mechanism, the separation between the slits S₃ and S₄ can be changed. The intensity at the point P which is at the common perpendicular bisector of S₁S₂ and S₃S₄. When z = D𝜆/2d, the intensity measured at P is I. Find the intensity when z is equal to
(a) D𝜆/d
(b) 3D𝜆/2d, and
(c) 2D𝜆/d
ANSWER: Let the intensity of S₁ and S₂ = I'. For S₃ and S₄, y = z/2. Hence the path difference here is
Δx = yd/D =zd/2D
=(D𝜆/2d)d/2D
=𝜆/4
The phase difference ẟ = 2π(Δx)/𝜆 =2π𝜆/4𝜆 =π/2.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π/4) =4I'*½ =2I'
where I' is the intensity of each of the sources. For the point P, the holes S₃ and S₄ are sources with intensity 2I' of each. Also, the point P is the point of central maxima, hence the intensity at P is
I = 4*2I' =8I'.
→I' = I/8.
(a) when z = D𝜆/d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (D𝜆/d)*d/2D =𝜆/2
The phase difference ẟ = 2π(Δx)/𝜆 =2π𝜆/2𝜆 =π.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π/2) =4I'*0 =0
So the intensities at each of the slits S₃ and S₄ is zero and will be dark. Therefore the intensity at the point P is also zero.
The phase difference ẟ = 2π(Δx)/𝜆 =2π(3𝜆/4)/𝜆 =6π/4 =3π/2.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(3π/4) =4I'*½ =2I'
For the point P, the slits S₃ and S₄ are sources with intensity 2I' each. Hence the intensity at the central point P where maxima forms is = 4*2I' = 8I' =I.
(c) when z = 2D𝜆/d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (2D𝜆/d)*d/2D =𝜆
The phase difference ẟ = 2π(Δx)/𝜆 =2π(𝜆)/𝜆 =2π
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π) =4I'*1 =4I'.
Thus the intensity at the maxima at P = 4*4I' =16I'
=16*(I/8) = 2I.
35. A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution, if it is known to be 1.2 and 1.5?
ANSWER: Given 𝜆 =580 nm =580x10⁻⁹ m
The thickness of the film, d =0.0011 mm = 1.1x10⁻⁶ m
Since the bubble appears dark when seen by the reflected light, destructive interference takes place, hence
2µd =n𝜆
→2µ =n*580x10⁻⁹/1.1x 10⁻⁶
→µ = n*0.264
Since it is known that µ is between 1.2 and 1.5, for the value of n = 5, µ = 5*0.264 =1.32.
36. A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index 1.4). What should be the minimum thickness of the film so that it strongly reflects the light?
ANSWER: Given 𝜆 =560 nm
µ = 1.4, Thickness d = ?
For the film to strongly reflect the light
2µd =(n+½)𝜆
→2*1.4*d = (n+½)*560
→d = (n+½)*200 nm
The minimum value of d will be for n = 0
→d = 200/2 =100 nm
37. A parallel beam of white light is incident normally on a water film 1.0x10⁻⁴ cm thick. Find the wavelength in the visible range (400 nm - 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33.
ANSWER: For the light to be strongly transmitted by the film,
2µd = n𝜆
→𝜆 = 2µd/n = 2*1.33*1.0x10⁻⁴*10⁹/100n
=2.66x1000/n =2660/n
For n = 4, 𝜆 = 665 nm
for n = 5, 𝜆 = 532 nm
for n = 6, 𝜆 = 443 nm
For other values of n, the 𝜆 is beyond the range 400 nm - 700 nm.
38. A glass surface is coated by an oil film of uniform thickness 1.0x10⁻⁴ cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelength of the light in the visible range (400 nm-750 nm) which are completely transmitted by the oil film under normal incidence.
ANSWER: For the oil thickness of the film, d = 1.0x10⁻⁴ cm, µ = 1.25.
For complete transmission under the normal incidence, 2µd = (n+½)𝜆.
→𝜆 = 2µd/(n+½)
=2*1.25*1.0x10⁻⁴*10⁷/(n+½)
=2.5x1000/(n+½) nm
=2500/(n+½) nm
For n = 3, 𝜆 = 2500/3.5 =714 nm,
n = 4, 𝜆 = 2500/4.5 =556 nm,
n = 5, 𝜆 = 2500/5.5 =455 nm.
For other values of n, 𝜆 is out of the given range.
39. Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30°.
ANSWER: The first diffraction minimum is given at
b.sinθ = 𝜆
→𝜆 = b.sinθ = 5.sin30° cm
=2.5 cm
40. Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and fall on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?
ANSWER: The bright spot is the circular area within the first dark ring. The radius of the central bright spot
R = 1.22𝜆D/b
Here 𝜆 = 560 nm =560x10⁻⁹ m
D = 2.0 m
b = 0.20 mm =0.0002 m
→R = 1.22*560x10⁻⁹*2/0.0002 m
=1.22*560x10⁻⁵*100 cm
≈ 0.68 cm
41. A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?
ANSWER: The distance D = 20 cm =0.20 m
𝜆 = 620 nm =620x10⁻⁹ m
The radius of the lens, b = 8/2 cm =4 cm =0.04 m
The radius of the central bright spot is given as
R = 1.22𝜆D/b
= 1.22*620x10⁻⁹*0.2/0.04 m
= 3782x10⁻⁹ m
≈ 3.8x10⁻⁶ m
ANSWER: If the intensity of each of the slits =I', the resultant intensity I =4I'*cos²(ẟ/2)
where ẟ is the phase difference.
→cos²(ẟ/2) = I/4I'
 |
| Diagram for Q-31 |
For the central maximum, the intensity at the bright fringe =4I', because of ẟ =2nπ. Suppose at a distance y from the central maximum the intensity is half of the maximum intensity. i.e. I = 2I'.
→cos²(ẟ/2) = 2I'/4I' =½
→cos(ẟ/2) = 1/√2
→ẟ/2 = π/4
→ẟ = π/2 = phase difference.
→2π*Δx/𝜆 = π/2
→Δx = 𝜆/4
→yd/D =𝜆/4
→y = 𝜆D/4d = 500x10⁻⁹ *1.0/(4*0.001)
→y = 1.25x10⁻⁴ m
32. The line width of a bright fringe is some times defined as the separation between the points on the two sides the central line where the intensity falls to half the maximum. Find the line width of a bright fringe in a Young's double slit experiment in terms of 𝜆, d and D where the symbols have their usual meaning.
ANSWER: If the intensity of each of the slits =I', the resultant intensity I =4I'*cos²(ẟ/2)
where ẟ is the phase difference.
→cos²(ẟ/2) = I/4I'
For the central maximum, the intensity at the bright fringe =4I', because of ẟ =2nπ. Suppose at a distance y from the central maximum the intensity is half of the maximum intensity. i.e. I = 2I'.
→cos²(ẟ/2) = 2I'/4I' =½
→cos(ẟ/2) = 1/√2
→ẟ/2 = π/4
→ẟ = π/2 = phase difference.
→2π*Δx/𝜆 = π/2
→Δx = 𝜆/4
→yd/D =𝜆/4
→y = 𝜆D/4d
But the line width is the separation between the points on either side of the central maximum where the intensity falls to half of the maximum. Hence the line width = 2y = 2𝜆D/4d = 𝜆D/2d
33. Consider the situation shown in figure (17-E6). The two slits S₁ and S₂ placed symmetrically around the central line are illuminated by monochromatic light of wavelength 𝜆. The separation between the slits is d. The light transmitted by the slits falls on a screen Σ₁ placed at a distance D from the slits. The slit S₃ is at the central line and the slit S₄ is at a distance z from S₃. Another screen Σ₂ is placed a further distance D away from Σ₁. Find the ratio of the maximum to minimum intensity observed on Σ₂ if z is equal to
(a) 𝜆D/2d
(b) 𝜆D/d
(c) 𝜆D/4d
 |
| Figure for Q-33 |
ANSWER: Let the intensity of each of the slits S₁ and S₂ =I'. The corresponding amplitude = r (say). S₃ is at the center of the screen where the path difference and hence the phase difference is zero. So here the intensity is maximum = I (say). Hence I = 4I'.
At S₄, the path difference Δx = zd/D
Phase difference ẟ = 2π*Δx/𝜆 =2πzd/𝜆D
(a) For z =𝜆D/2d
ẟ = (2πd/𝜆D)*(𝜆D/2d) =π
Since at S₄, the phase difference is π, the destructive interference takes place and the minima occur where the intensity is zero.
So for the screen Σ₂, out of the two slits S₃ and S₄, only S₃ is illuminated with an intensity I. The screen will be uniformly illuminated having the same value for maxima and minima. Hence their ratio = 1
(b) For z =𝜆D/d
ẟ = (2πd/𝜆D)*(𝜆D/d) =2π
Since at S₄, the phase difference is 2π, the constructive interference takes place and the maxima occur where the intensity is I same as S₃.
So for the screen Σ₂, the two slits S₃ and S₄ are illuminated with an intensity I. They will form an interference pattern with bright and dark lines i.e. with maximum intensity = I" (say) and minimum intensity zero. Hence the ratio = I"/0 = ∞.
(c) For z =𝜆D/4d
ẟ = (2πd/𝜆D)*(𝜆D/4d) =π/2
Since at S₄, the phase difference = π/2, the intensity here is given as
I₁ = 4I'.cos²(ẟ/2) =4I'.cos²(π/4) =4I'*½ =2I'.
So S₃ is illuminated with an intensity I = 4I' and S₄ with intensity 2I'.
Let the amplitude for I' = r, for 2I' = r₁ and for 4I' =r₂. So,
2I'/I' = r₁²/r²
→r₁²/r² =2
→r₁² = 2r²
→r₁ =√2.r
And, 4I'/I' = r₂²/r²
→r₂²/r² = 4
→r₂² = 4r²
→r₂ = 2r
Hence the amplitude of the light wave coming from S₃ is 2r and that coming from S₄ is √2.r, so when they interfere the amplitude of the maximum intensity (say Iₘ) =2r+√2.r and the amplitude of the minimum intensity (say Iₙ) = 2r-√2r. Thus the ratio of maximum to minimum intensity on the screen Σ₂
Iₘ/Iₙ = (2r+√2.r)²/(2r-√2.r)²
=(2+√2)²/(2-√2)²
=(4+2+4√2)/(4+2-4√2)
=(6+4√2)/(6-4√2)
=(3+2√2)/(3-2√2)
Multiply numerator and denominator by (3+2√2)
→Iₘ/Iₙ = (3+2√2)²/(3²-8) =(9+8+12√2)/(9-8)
=(17+12√2)/1 =17+12*1.41 =17+16.92 ≈34
34. Consider the arrangement shown in figure (17-E7). By some mechanism, the separation between the slits S₃ and S₄ can be changed. The intensity at the point P which is at the common perpendicular bisector of S₁S₂ and S₃S₄. When z = D𝜆/2d, the intensity measured at P is I. Find the intensity when z is equal to
(a) D𝜆/d
(b) 3D𝜆/2d, and
(c) 2D𝜆/d
 |
| Figure for Q-34 |
ANSWER: Let the intensity of S₁ and S₂ = I'. For S₃ and S₄, y = z/2. Hence the path difference here is
Δx = yd/D =zd/2D
=(D𝜆/2d)d/2D
=𝜆/4
The phase difference ẟ = 2π(Δx)/𝜆 =2π𝜆/4𝜆 =π/2.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π/4) =4I'*½ =2I'
where I' is the intensity of each of the sources. For the point P, the holes S₃ and S₄ are sources with intensity 2I' of each. Also, the point P is the point of central maxima, hence the intensity at P is
I = 4*2I' =8I'.
→I' = I/8.
(a) when z = D𝜆/d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (D𝜆/d)*d/2D =𝜆/2
The phase difference ẟ = 2π(Δx)/𝜆 =2π𝜆/2𝜆 =π.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π/2) =4I'*0 =0
So the intensities at each of the slits S₃ and S₄ is zero and will be dark. Therefore the intensity at the point P is also zero.
(b) when z = 3D𝜆/2d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (3D𝜆/2d)*d/2D =3𝜆/4The phase difference ẟ = 2π(Δx)/𝜆 =2π(3𝜆/4)/𝜆 =6π/4 =3π/2.
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(3π/4) =4I'*½ =2I'
For the point P, the slits S₃ and S₄ are sources with intensity 2I' each. Hence the intensity at the central point P where maxima forms is = 4*2I' = 8I' =I.
(c) when z = 2D𝜆/d, the path difference at S₃ and S₄ is,
Δx = yd/D =zd/2D = (2D𝜆/d)*d/2D =𝜆
The phase difference ẟ = 2π(Δx)/𝜆 =2π(𝜆)/𝜆 =2π
Hence intensities at each of the slits
I'' =4I'*cos²(ẟ/2) =4I'*cos²(π) =4I'*1 =4I'.
Thus the intensity at the maxima at P = 4*4I' =16I'
=16*(I/8) = 2I.
35. A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution, if it is known to be 1.2 and 1.5?
ANSWER: Given 𝜆 =580 nm =580x10⁻⁹ m
The thickness of the film, d =0.0011 mm = 1.1x10⁻⁶ m
Since the bubble appears dark when seen by the reflected light, destructive interference takes place, hence
2µd =n𝜆
→2µ =n*580x10⁻⁹/1.1x 10⁻⁶
→µ = n*0.264
Since it is known that µ is between 1.2 and 1.5, for the value of n = 5, µ = 5*0.264 =1.32.
36. A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index 1.4). What should be the minimum thickness of the film so that it strongly reflects the light?
ANSWER: Given 𝜆 =560 nm
µ = 1.4, Thickness d = ?
For the film to strongly reflect the light
2µd =(n+½)𝜆
→2*1.4*d = (n+½)*560
→d = (n+½)*200 nm
The minimum value of d will be for n = 0
→d = 200/2 =100 nm
37. A parallel beam of white light is incident normally on a water film 1.0x10⁻⁴ cm thick. Find the wavelength in the visible range (400 nm - 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33.
ANSWER: For the light to be strongly transmitted by the film,
2µd = n𝜆
→𝜆 = 2µd/n = 2*1.33*1.0x10⁻⁴*10⁹/100n
=2.66x1000/n =2660/n
For n = 4, 𝜆 = 665 nm
for n = 5, 𝜆 = 532 nm
for n = 6, 𝜆 = 443 nm
For other values of n, the 𝜆 is beyond the range 400 nm - 700 nm.
38. A glass surface is coated by an oil film of uniform thickness 1.0x10⁻⁴ cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelength of the light in the visible range (400 nm-750 nm) which are completely transmitted by the oil film under normal incidence.
ANSWER: For the oil thickness of the film, d = 1.0x10⁻⁴ cm, µ = 1.25.
For complete transmission under the normal incidence, 2µd = (n+½)𝜆.
→𝜆 = 2µd/(n+½)
=2*1.25*1.0x10⁻⁴*10⁷/(n+½)
=2.5x1000/(n+½) nm
=2500/(n+½) nm
For n = 3, 𝜆 = 2500/3.5 =714 nm,
n = 4, 𝜆 = 2500/4.5 =556 nm,
n = 5, 𝜆 = 2500/5.5 =455 nm.
For other values of n, 𝜆 is out of the given range.
39. Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30°.
ANSWER: The first diffraction minimum is given at
b.sinθ = 𝜆
→𝜆 = b.sinθ = 5.sin30° cm
=2.5 cm
40. Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and fall on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?
ANSWER: The bright spot is the circular area within the first dark ring. The radius of the central bright spot
R = 1.22𝜆D/b
Here 𝜆 = 560 nm =560x10⁻⁹ m
D = 2.0 m
b = 0.20 mm =0.0002 m
→R = 1.22*560x10⁻⁹*2/0.0002 m
=1.22*560x10⁻⁵*100 cm
≈ 0.68 cm
41. A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?
ANSWER: The distance D = 20 cm =0.20 m
𝜆 = 620 nm =620x10⁻⁹ m
The radius of the lens, b = 8/2 cm =4 cm =0.04 m
The radius of the central bright spot is given as
R = 1.22𝜆D/b
= 1.22*620x10⁻⁹*0.2/0.04 m
= 3782x10⁻⁹ m
≈ 3.8x10⁻⁶ m
===<<<O>>>===
Links to the Chapters
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CHAPTER- 16 - Sound Waves
OBJECTIVE-I
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EXERCISES - Q-1 TO Q-10
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EXERCISES -Q-31 TO Q-40
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EXERCISES -Q-51 TO Q-60
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CHAPTER- 15 - Wave Motion and Waves on a String
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EXERCISES - Q-1 TO Q-10
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 11 - Gravitation
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CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 11 - Gravitation
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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