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LIGHT WAVES
EXERCISES Q-21 TO Q-30
21. A narrow slit S transmitting light of wavelength 𝜆 is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen Σ placed at a distance D from the slit.
(a) what will be intensity at a point just above the mirror, i.e. just above O?
(b) At what distance from O does the first maximum occur?
Figure for Q-21
ANSWER: (a) The light wave just after the reflection suffers a sudden phase-change of π. Just above the mirror direct and reflected waves interfere. Though there is no geometrical path difference between the two waves but they have a phase difference of π, Which is a condition of destructive interference. The resultant amplitude of light will be zero hence the resultant intensity is zero.
(b) The maximum will occur where the path difference between the two waves is (n+½)𝜆. It is just opposite the normal case because here the reflected wave suffers a sudden phase difference of π. The path difference can be calculated assuming the reflected wave coming from a virtual slit which will be the image of the real slit. Hence the slit separation =2d.
The path difference at a point on the screen y distance above the center Δx = (2d)*y/D
Hence for maxima (2d)*y/D = (n+½)𝜆
For the first maxima, n = 0, hence
(2d)*y/D = ½𝜆
→y = 𝜆D/4d
22. A long narrow horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. Find the fringe width if the light used has a wavelength of 700 nm.
ANSWER: The interference pattern on the screen will be due to the direct wave and the reflected wave. The reflected wave will be coming from a virtual slit which will be the image of the original slit in the mirror. Hence the slit separation d = 2*distance of slit above the mirror
→d = 2*1 mm = 2 mm = 0.002 m
D = 1.0 m
𝜆 = 700 nm = 700* 10⁻⁹ m = 7x10⁻⁷ m
Hence the fringe width, w = D𝜆/d
→w = 1.0*7x10⁻⁷/0.002 m =3.5x10⁻⁴ m
→w = 3.5x10⁻⁴*1000 mm = 0.35 mm
23. Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?
ANSWER: Let the intensity of the direct light wave = I, then the intensity of the reflected light I' = 0.64I. Hence the ratio of the intensities,
I/I' = I/0.64I =100/64
But the intensities are proportional to the square of the amplitudes. If the amplitude of the direct wave is r and that of reflected wave =r', then
I/I' =r²/r'² = 100/64
→r/r' = 10/8 =5/4
→r/5 = r'/4 = k (say)
→r = 5k and r' = 4k
Hence on interference maximum amplitude, Rm =5k+4k =9k and the minimum amplitude Rn =5k-4k =k.
If the intensities at maximum and the minimum be Im and In respectively, then
Im/In = Rm²/Rn² =(9k)²/(k)² = 81:1
24. A double slit S₁ - S₂ is illuminated by coherent light of wavelength 𝜆. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D₁ from it and a screen Σ is placed behind the double slit at a distance D₂ from it (figure 17-E2). The screen Σ receives only the light reflected by the mirror. Find the fringe width of the interference pattern on the screen.
Figure for Q-24
ANSWER: Since both of the interfering waves are reflected hence both have a sudden phase change of π. The waves will interfere normally. The distance of the screen, in this case, will be the distance of the virtual screen (image of the screen) from the slits D. D = (D₁+D₂)+D₁ =2D₁+D₂
Hence the fringe width of the interference pattern,
w = 𝜆D/d =𝜆(2D₁+D₂)/d.
25. White coherent light (400 nm - 700 nm) is sent through the slits of a Young's double slit experiment (figure 17E-3). The separation between the slits is 0.5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringes) from the central line.
(a) which wavelength(s) will be absent in the light coming from the hole?
(b) which wavelength will have a strong intensity?
Figure for Q-25
ANSWER: (a) The path difference at a point y distance above the center of the screen,
Δx = d*y/D
And for the minima the path difference
Δx = (n+½)𝜆
Hence
dy/D =(n+½)𝜆
→𝜆 = dy/{D(n+½)}
Here y = 1 mm =0.001 m
d = 0.5 mm =0.0005 m
D = 0.50 m
→𝜆 = 0.0005*0.001/{0.50(n+½)}
→𝜆 = 1x10⁻⁶/(n+½) m
→𝜆 = 1x10⁻⁶*10⁹/(n+½) nm =1x10³/(n+½) nm
The range of white light is 400 - 700 nm.
If we put n = 0, the 𝜆 = 2000 nm which is out of the range.
If we put n = 1, 𝜆 = 2000/3 nm =667 nm, (within the range)
For n = 2, 𝜆 = 2000/5 =400 nm (within the range)
For n = 3, 𝜆 = 2000/7 =286 nm (out of the range).
It is clear that the wavelengths of 400 nm and 667 nm will be absent from the light coming out of the hole.
(b) The strong intensity will be at maxima and for the maxima path difference Δx = n𝜆
→dy/D = n𝜆
→𝜆 = dy/Dn
→𝜆 = 0.0005*0.001/0.5n
→𝜆 = 1x10⁻⁶/n m
→𝜆 =1x10⁻⁶*10⁹/n nm =1000/n nm
for n = 1, 𝜆 = 1000 nm but it is out of range.
for n = 2, 𝜆 = 500 nm which is within the range.
for n = 3, 𝜆 = 333 nm but is out of range.
Hence the maximum intensity at the hole will be for the wavelength 500 nm.
26. Consider the arrangement shown in the figure (17E-4). The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe width.
Figure for Q-26
ANSWER: (a) There will be a dark fringe at O if the path difference between the waves coming here from the two slits is equal to (n+½)𝜆. Let us draw a diagram as below.
Diagram for Q-26
Path difference Δx = ACO - ABO =2(AC-AB) =2{√(D²+d²)-D}
Hence for minimum at O,
2{√(D²+d²)-D} = (n+½)𝜆
→√(D²+d²)-D = (n+½)𝜆/2
→√(D²+d²) = D + (n+½)𝜆/2
Square both sides.
D²+d² = D² +D(n+½)𝜆 + (n+½)²𝜆²/4
Since (n+½)²𝜆²/4 is very very small in comparison to others (due to 𝜆²), we neglect it. Thus
D²+d² = D² +D(n+½)𝜆
→d² = D(n+½)𝜆
→d =√{D(n+½)𝜆}
For d to be minimum, n should be minimum i.e. n = 0.
→d = √(𝜆D/2)
(b) Let us draw a diagram.
Diagram for Q-26(b)
At P a distance x above O, the path difference is
Δx =AC+CP-AB-BP
=(AC-AB)+(CP-BP)
Let us see first AC-AB. Considering a line parallel to ABO passing through the middle of BC.
We know that Δx =yd/D. Here y = d/2, hence
AC-AB =d²/2D
Similarly it can be shown that CP-BP ≈ (d/2-x)d/D
Since y = d/2-x
So, for maximum
d²/2D + (d/2-x)d/D =n𝜆
→d²/2D+d²/2D -xd/D =n𝜆
→d²/D -n𝜆 =xd/D
→x = d-n𝜆D/d =d -n*2d
[Since 𝜆D/2 = d², 𝜆D =2d²]
For next bright fringe n = 1
→x = d-2d =-d
So the next bright fringe will be at a distance d from O.
(c) The fringe width =2*OP =2d
27. The coherent point source S₁ and S₂ vibrating in phase emit light of wavelength 𝜆. The separation between the sources is 2𝜆. Consider a line passing through S₂ and perpendicular to the line S₁S₂. What is the smallest distance from S₂ where a minimum of intensity occurs?
ANSWER: Let there be a point P on the line S₂P such that S₂P =x. Given S₁S₂ =2𝜆. The path difference of the two waves reaching P =S₁P-S₂P
=√{x²+(2𝜆)²}-x
Diagram for Q-27
For a minimum,
√{x²+(2𝜆)²}-x = (n+½)𝜆
→√{x²+(2𝜆)²} = x+(n+½)𝜆
→x²+4𝜆² =x²+(n+½)²𝜆²+2x(n+½)𝜆
→2x(n+½) = 4𝜆-(n+½)²𝜆
→x = {4-(n+½)²}𝜆/2(n+½)
For n = 0, x = 15𝜆/4
for n = 1, x = {4-9/4}𝜆/3 =7𝜆/12
for n = 2, x = -9𝜆/20 {negative value not acceptable}
Hence the minimum value of x for a minimum intensity = 7𝜆/12.
28. Figure (17E-5) three equidistant slits being illuminated by a monochromatic parallel beam of light. Let BP₀-AP₀ =𝜆/3 and D>>𝜆.
(a) Show that in this case d = √(2𝜆D/3).
(b) Show that intensity at P₀ is three times the intensity due to any of the three slits individually.
Figure for Q-28
ANSWER: (a) Considering a horizontal reference line through the middle of AB. The distance of P₀ from this reference line =y =d/2. In this case, the path difference between BP₀ and AP₀ is
BP₀-AP₀ =y*d/D
→𝜆/3 = (d/2)*(d/D) =d²/2D
→d² = 2𝜆D/3
→d = √(2𝜆D/3)
Proved.
(b) Considering the horizontal reference line through the middle of BC, the distance of P₀ from it = d+d/2 =3d/2.
Hence the path difference between the waves coming from B and C = (3d/2)*d/D =3d²/2D =3*(2𝜆D/3)/2D =6𝜆D/6D =𝜆. So both are in the same phase.
Consider all the three waves coming at P₀. The waves from B and C are in the same phase while the wave coming from A has a phase difference
=2π(𝜆/3)/𝜆 =2π/3.
Let the intensity from each of the slits = I and corresponding maximum amplitude = r.
Diagram for Q-28
The resultant amplitude will be the vector sum of the individual amplitudes. Since the waves from B and C are in the same phase sum of both amplitudes = 2r. The amplitude of the wave coming from A makes an angle 2π/3 from the other two. Hence the resultant of 2r and r having an angle 2π/3 is given as
R² =(2r)²+r²+2.2r.r.cos(2π/3)
=5r²+4r²(-½) =5r²-2r² =3r²
Since the intensity is proportional to the amplitude of the wave. Hence the ratio of total intensity to one of the individual intensity = I'/I =R²/r²
→I'/I = 3r²/r² = 3
→I' = 3I
So, the intensity at P₀ is three times the intensity due to any of the three slits individually
29. In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the center of the central maximum is 0.20 W/m², what will be the intensity at a point 0.50 cm away from this center along the width of the fringes?
ANSWER: Given d = 2.0 mm = 0.002 m, D = 2.0 m, 𝜆 = 600*10⁻⁹ m =6x10⁻⁷ m.
If the intensity of each of the slits = I and the corresponding maximum amplitude of the waves = r, then amplitude at the center of the central maximum = 2r, because both the waves are in phase here. If the resultant intensity is I' then
I'/I = (2r)²/r² =4
→I = I'/4
The point in the question is at y = 0.50 cm =0.005 m. Hence the path difference between the waves
Δx = yd/D =0.005*0.002/2 m =5x10⁻⁶ m
Phase difference =2π*Δx/𝜆
=2π*5x10⁻⁶/6x10⁻⁷ =50π/3 =16π+2π/3→2π/3
The resultant amplitude here
R² = r²+r²+2.r.r.cos(2π/3) =2r²+2r²(-½) =2r²-r² =r²
→R = r
It is the amplitude corresponding to the individual intensity I. Hence the intensity at this point
I = I'/4 = 0.20/4 W/m² =0.05 W/m²
30. In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength 𝜆. Find the distance from the central point where the intensity falls to
(a) half the maximum,
(b) one-fourth of the maximum.
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'
(a) The maximum intensity at the bright fringe =4I', because of ẟ =2nπ.
The intensity half of the maximum = 2I', here
cos²(ẟ/2) =2I'/4I' =½
→cos(ẟ/2) =1/√2
→ẟ/2 =π/4
→ẟ = π/2
But Phase difference ẟ = 2π*Δx/𝜆
→Δx = ẟ𝜆/2π
So for the intensity half of the maximum, the phase difference will be π/2. And the path difference
Δx = π𝜆/4π =𝜆/4
If this point is at a distance y from the central point,
The path difference Δx =yd/D
→y =D*Δx/d =D*𝜆/4d
→y = D𝜆/4d
(b) The intensity one-fourth of the maximum =4I'/4 =I'
Hence cos²(ẟ/2) =I'/4I' =1/4
→cos(ẟ/2) = 1/2 =cos(π/3)
→ẟ/2 = π/3
→ẟ = 2π/3
Thus Δx = ẟ𝜆/2π =2π𝜆/6π =𝜆/3
Since Δx =yd/D
→𝜆/3 =yd/D
→y = D𝜆/3d
21. A narrow slit S transmitting light of wavelength 𝜆 is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen Σ placed at a distance D from the slit.
(a) what will be intensity at a point just above the mirror, i.e. just above O?
(b) At what distance from O does the first maximum occur?
ANSWER: (a) The light wave just after the reflection suffers a sudden phase-change of π. Just above the mirror direct and reflected waves interfere. Though there is no geometrical path difference between the two waves but they have a phase difference of π, Which is a condition of destructive interference. The resultant amplitude of light will be zero hence the resultant intensity is zero.
(b) The maximum will occur where the path difference between the two waves is (n+½)𝜆. It is just opposite the normal case because here the reflected wave suffers a sudden phase difference of π. The path difference can be calculated assuming the reflected wave coming from a virtual slit which will be the image of the real slit. Hence the slit separation =2d.
The path difference at a point on the screen y distance above the center Δx = (2d)*y/D
Hence for maxima (2d)*y/D = (n+½)𝜆
For the first maxima, n = 0, hence
(2d)*y/D = ½𝜆
→y = 𝜆D/4d
22. A long narrow horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. Find the fringe width if the light used has a wavelength of 700 nm.
ANSWER: The interference pattern on the screen will be due to the direct wave and the reflected wave. The reflected wave will be coming from a virtual slit which will be the image of the original slit in the mirror. Hence the slit separation d = 2*distance of slit above the mirror
→d = 2*1 mm = 2 mm = 0.002 m
D = 1.0 m
𝜆 = 700 nm = 700* 10⁻⁹ m = 7x10⁻⁷ m
Hence the fringe width, w = D𝜆/d
→w = 1.0*7x10⁻⁷/0.002 m =3.5x10⁻⁴ m
→w = 3.5x10⁻⁴*1000 mm = 0.35 mm
23. Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?
ANSWER: Let the intensity of the direct light wave = I, then the intensity of the reflected light I' = 0.64I. Hence the ratio of the intensities,
I/I' = I/0.64I =100/64
But the intensities are proportional to the square of the amplitudes. If the amplitude of the direct wave is r and that of reflected wave =r', then
I/I' =r²/r'² = 100/64
→r/r' = 10/8 =5/4
→r/5 = r'/4 = k (say)
→r = 5k and r' = 4k
Hence on interference maximum amplitude, Rm =5k+4k =9k and the minimum amplitude Rn =5k-4k =k.
If the intensities at maximum and the minimum be Im and In respectively, then
Im/In = Rm²/Rn² =(9k)²/(k)² = 81:1
24. A double slit S₁ - S₂ is illuminated by coherent light of wavelength 𝜆. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D₁ from it and a screen Σ is placed behind the double slit at a distance D₂ from it (figure 17-E2). The screen Σ receives only the light reflected by the mirror. Find the fringe width of the interference pattern on the screen.
ANSWER: Since both of the interfering waves are reflected hence both have a sudden phase change of π. The waves will interfere normally. The distance of the screen, in this case, will be the distance of the virtual screen (image of the screen) from the slits D. D = (D₁+D₂)+D₁ =2D₁+D₂
Hence the fringe width of the interference pattern,
w = 𝜆D/d =𝜆(2D₁+D₂)/d.
25. White coherent light (400 nm - 700 nm) is sent through the slits of a Young's double slit experiment (figure 17E-3). The separation between the slits is 0.5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringes) from the central line.
(a) which wavelength(s) will be absent in the light coming from the hole?
(b) which wavelength will have a strong intensity?
ANSWER: (a) The path difference at a point y distance above the center of the screen,
Δx = d*y/D
And for the minima the path difference
Δx = (n+½)𝜆
Hence
dy/D =(n+½)𝜆
→𝜆 = dy/{D(n+½)}
Here y = 1 mm =0.001 m
d = 0.5 mm =0.0005 m
D = 0.50 m
→𝜆 = 0.0005*0.001/{0.50(n+½)}
→𝜆 = 1x10⁻⁶/(n+½) m
→𝜆 = 1x10⁻⁶*10⁹/(n+½) nm =1x10³/(n+½) nm
The range of white light is 400 - 700 nm.
If we put n = 0, the 𝜆 = 2000 nm which is out of the range.
If we put n = 1, 𝜆 = 2000/3 nm =667 nm, (within the range)
For n = 2, 𝜆 = 2000/5 =400 nm (within the range)
For n = 3, 𝜆 = 2000/7 =286 nm (out of the range).
It is clear that the wavelengths of 400 nm and 667 nm will be absent from the light coming out of the hole.
(b) The strong intensity will be at maxima and for the maxima path difference Δx = n𝜆
→dy/D = n𝜆
→𝜆 = dy/Dn
→𝜆 = 0.0005*0.001/0.5n
→𝜆 = 1x10⁻⁶/n m
→𝜆 =1x10⁻⁶*10⁹/n nm =1000/n nm
for n = 1, 𝜆 = 1000 nm but it is out of range.
for n = 2, 𝜆 = 500 nm which is within the range.
for n = 3, 𝜆 = 333 nm but is out of range.
Hence the maximum intensity at the hole will be for the wavelength 500 nm.
26. Consider the arrangement shown in the figure (17E-4). The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe width.
ANSWER: (a) There will be a dark fringe at O if the path difference between the waves coming here from the two slits is equal to (n+½)𝜆. Let us draw a diagram as below.
Path difference Δx = ACO - ABO =2(AC-AB) =2{√(D²+d²)-D}
Hence for minimum at O,
2{√(D²+d²)-D} = (n+½)𝜆
→√(D²+d²)-D = (n+½)𝜆/2
→√(D²+d²) = D + (n+½)𝜆/2
Square both sides.
D²+d² = D² +D(n+½)𝜆 + (n+½)²𝜆²/4
Since (n+½)²𝜆²/4 is very very small in comparison to others (due to 𝜆²), we neglect it. Thus
D²+d² = D² +D(n+½)𝜆
→d² = D(n+½)𝜆
→d =√{D(n+½)𝜆}
For d to be minimum, n should be minimum i.e. n = 0.
→d = √(𝜆D/2)
(b) Let us draw a diagram.
At P a distance x above O, the path difference is
Δx =AC+CP-AB-BP
=(AC-AB)+(CP-BP)
Let us see first AC-AB. Considering a line parallel to ABO passing through the middle of BC.
We know that Δx =yd/D. Here y = d/2, hence
AC-AB =d²/2D
Similarly it can be shown that CP-BP ≈ (d/2-x)d/D
Since y = d/2-x
So, for maximum
d²/2D + (d/2-x)d/D =n𝜆
→d²/2D+d²/2D -xd/D =n𝜆
→d²/D -n𝜆 =xd/D
→x = d-n𝜆D/d =d -n*2d
[Since 𝜆D/2 = d², 𝜆D =2d²]
For next bright fringe n = 1
→x = d-2d =-d
So the next bright fringe will be at a distance d from O.
(c) The fringe width =2*OP =2d
27. The coherent point source S₁ and S₂ vibrating in phase emit light of wavelength 𝜆. The separation between the sources is 2𝜆. Consider a line passing through S₂ and perpendicular to the line S₁S₂. What is the smallest distance from S₂ where a minimum of intensity occurs?
ANSWER: Let there be a point P on the line S₂P such that S₂P =x. Given S₁S₂ =2𝜆. The path difference of the two waves reaching P =S₁P-S₂P
=√{x²+(2𝜆)²}-x
For a minimum,
√{x²+(2𝜆)²}-x = (n+½)𝜆
→√{x²+(2𝜆)²} = x+(n+½)𝜆
→x²+4𝜆² =x²+(n+½)²𝜆²+2x(n+½)𝜆
→2x(n+½) = 4𝜆-(n+½)²𝜆
→x = {4-(n+½)²}𝜆/2(n+½)
For n = 0, x = 15𝜆/4
for n = 1, x = {4-9/4}𝜆/3 =7𝜆/12
for n = 2, x = -9𝜆/20 {negative value not acceptable}
Hence the minimum value of x for a minimum intensity = 7𝜆/12.
28. Figure (17E-5) three equidistant slits being illuminated by a monochromatic parallel beam of light. Let BP₀-AP₀ =𝜆/3 and D>>𝜆.
(a) Show that in this case d = √(2𝜆D/3).
(b) Show that intensity at P₀ is three times the intensity due to any of the three slits individually.
ANSWER: (a) Considering a horizontal reference line through the middle of AB. The distance of P₀ from this reference line =y =d/2. In this case, the path difference between BP₀ and AP₀ is
BP₀-AP₀ =y*d/D
→𝜆/3 = (d/2)*(d/D) =d²/2D
→d² = 2𝜆D/3
→d = √(2𝜆D/3)
Proved.
(b) Considering the horizontal reference line through the middle of BC, the distance of P₀ from it = d+d/2 =3d/2.
Hence the path difference between the waves coming from B and C = (3d/2)*d/D =3d²/2D =3*(2𝜆D/3)/2D =6𝜆D/6D =𝜆. So both are in the same phase.
Consider all the three waves coming at P₀. The waves from B and C are in the same phase while the wave coming from A has a phase difference
=2π(𝜆/3)/𝜆 =2π/3.
Let the intensity from each of the slits = I and corresponding maximum amplitude = r.
The resultant amplitude will be the vector sum of the individual amplitudes. Since the waves from B and C are in the same phase sum of both amplitudes = 2r. The amplitude of the wave coming from A makes an angle 2π/3 from the other two. Hence the resultant of 2r and r having an angle 2π/3 is given as
R² =(2r)²+r²+2.2r.r.cos(2π/3)
=5r²+4r²(-½) =5r²-2r² =3r²
Since the intensity is proportional to the amplitude of the wave. Hence the ratio of total intensity to one of the individual intensity = I'/I =R²/r²
→I'/I = 3r²/r² = 3
→I' = 3I
So, the intensity at P₀ is three times the intensity due to any of the three slits individually
29. In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the center of the central maximum is 0.20 W/m², what will be the intensity at a point 0.50 cm away from this center along the width of the fringes?
ANSWER: Given d = 2.0 mm = 0.002 m, D = 2.0 m, 𝜆 = 600*10⁻⁹ m =6x10⁻⁷ m.
If the intensity of each of the slits = I and the corresponding maximum amplitude of the waves = r, then amplitude at the center of the central maximum = 2r, because both the waves are in phase here. If the resultant intensity is I' then
I'/I = (2r)²/r² =4
→I = I'/4
The point in the question is at y = 0.50 cm =0.005 m. Hence the path difference between the waves
Δx = yd/D =0.005*0.002/2 m =5x10⁻⁶ m
Phase difference =2π*Δx/𝜆
=2π*5x10⁻⁶/6x10⁻⁷ =50π/3 =16π+2π/3→2π/3
The resultant amplitude here
R² = r²+r²+2.r.r.cos(2π/3) =2r²+2r²(-½) =2r²-r² =r²
→R = r
It is the amplitude corresponding to the individual intensity I. Hence the intensity at this point
I = I'/4 = 0.20/4 W/m² =0.05 W/m²
30. In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength 𝜆. Find the distance from the central point where the intensity falls to
(a) half the maximum,
(b) one-fourth of the maximum.
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'
(a) The maximum intensity at the bright fringe =4I', because of ẟ =2nπ.
The intensity half of the maximum = 2I', here
cos²(ẟ/2) =2I'/4I' =½
→cos(ẟ/2) =1/√2
→ẟ/2 =π/4
→ẟ = π/2
But Phase difference ẟ = 2π*Δx/𝜆
→Δx = ẟ𝜆/2π
So for the intensity half of the maximum, the phase difference will be π/2. And the path difference
Δx = π𝜆/4π =𝜆/4
If this point is at a distance y from the central point,
The path difference Δx =yd/D
→y =D*Δx/d =D*𝜆/4d
→y = D𝜆/4d
(b) The intensity one-fourth of the maximum =4I'/4 =I'
Hence cos²(ẟ/2) =I'/4I' =1/4
→cos(ẟ/2) = 1/2 =cos(π/3)
→ẟ/2 = π/3
→ẟ = 2π/3
Thus Δx = ẟ𝜆/2π =2π𝜆/6π =𝜆/3
Since Δx =yd/D
→𝜆/3 =yd/D
→y = D𝜆/3d
(a) what will be intensity at a point just above the mirror, i.e. just above O?
(b) At what distance from O does the first maximum occur?
 |
| Figure for Q-21 |
ANSWER: (a) The light wave just after the reflection suffers a sudden phase-change of π. Just above the mirror direct and reflected waves interfere. Though there is no geometrical path difference between the two waves but they have a phase difference of π, Which is a condition of destructive interference. The resultant amplitude of light will be zero hence the resultant intensity is zero.
(b) The maximum will occur where the path difference between the two waves is (n+½)𝜆. It is just opposite the normal case because here the reflected wave suffers a sudden phase difference of π. The path difference can be calculated assuming the reflected wave coming from a virtual slit which will be the image of the real slit. Hence the slit separation =2d.
The path difference at a point on the screen y distance above the center Δx = (2d)*y/D
Hence for maxima (2d)*y/D = (n+½)𝜆
For the first maxima, n = 0, hence
(2d)*y/D = ½𝜆
→y = 𝜆D/4d
22. A long narrow horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. Find the fringe width if the light used has a wavelength of 700 nm.
ANSWER: The interference pattern on the screen will be due to the direct wave and the reflected wave. The reflected wave will be coming from a virtual slit which will be the image of the original slit in the mirror. Hence the slit separation d = 2*distance of slit above the mirror
→d = 2*1 mm = 2 mm = 0.002 m
D = 1.0 m
𝜆 = 700 nm = 700* 10⁻⁹ m = 7x10⁻⁷ m
Hence the fringe width, w = D𝜆/d
→w = 1.0*7x10⁻⁷/0.002 m =3.5x10⁻⁴ m
→w = 3.5x10⁻⁴*1000 mm = 0.35 mm
23. Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?
ANSWER: Let the intensity of the direct light wave = I, then the intensity of the reflected light I' = 0.64I. Hence the ratio of the intensities,
I/I' = I/0.64I =100/64
But the intensities are proportional to the square of the amplitudes. If the amplitude of the direct wave is r and that of reflected wave =r', then
I/I' =r²/r'² = 100/64
→r/r' = 10/8 =5/4
→r/5 = r'/4 = k (say)
→r = 5k and r' = 4k
Hence on interference maximum amplitude, Rm =5k+4k =9k and the minimum amplitude Rn =5k-4k =k.
If the intensities at maximum and the minimum be Im and In respectively, then
Im/In = Rm²/Rn² =(9k)²/(k)² = 81:1
24. A double slit S₁ - S₂ is illuminated by coherent light of wavelength 𝜆. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D₁ from it and a screen Σ is placed behind the double slit at a distance D₂ from it (figure 17-E2). The screen Σ receives only the light reflected by the mirror. Find the fringe width of the interference pattern on the screen.
 |
| Figure for Q-24 |
ANSWER: Since both of the interfering waves are reflected hence both have a sudden phase change of π. The waves will interfere normally. The distance of the screen, in this case, will be the distance of the virtual screen (image of the screen) from the slits D. D = (D₁+D₂)+D₁ =2D₁+D₂
Hence the fringe width of the interference pattern,
w = 𝜆D/d =𝜆(2D₁+D₂)/d.
25. White coherent light (400 nm - 700 nm) is sent through the slits of a Young's double slit experiment (figure 17E-3). The separation between the slits is 0.5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringes) from the central line.
(a) which wavelength(s) will be absent in the light coming from the hole?
(b) which wavelength will have a strong intensity?
 |
| Figure for Q-25 |
ANSWER: (a) The path difference at a point y distance above the center of the screen,
Δx = d*y/D
And for the minima the path difference
Δx = (n+½)𝜆
Hence
dy/D =(n+½)𝜆
→𝜆 = dy/{D(n+½)}
Here y = 1 mm =0.001 m
d = 0.5 mm =0.0005 m
D = 0.50 m
→𝜆 = 0.0005*0.001/{0.50(n+½)}
→𝜆 = 1x10⁻⁶/(n+½) m
→𝜆 = 1x10⁻⁶*10⁹/(n+½) nm =1x10³/(n+½) nm
The range of white light is 400 - 700 nm.
If we put n = 0, the 𝜆 = 2000 nm which is out of the range.
If we put n = 1, 𝜆 = 2000/3 nm =667 nm, (within the range)
For n = 2, 𝜆 = 2000/5 =400 nm (within the range)
For n = 3, 𝜆 = 2000/7 =286 nm (out of the range).
It is clear that the wavelengths of 400 nm and 667 nm will be absent from the light coming out of the hole.
(b) The strong intensity will be at maxima and for the maxima path difference Δx = n𝜆
→dy/D = n𝜆
→𝜆 = dy/Dn
→𝜆 = 0.0005*0.001/0.5n
→𝜆 = 1x10⁻⁶/n m
→𝜆 =1x10⁻⁶*10⁹/n nm =1000/n nm
for n = 1, 𝜆 = 1000 nm but it is out of range.
for n = 2, 𝜆 = 500 nm which is within the range.
for n = 3, 𝜆 = 333 nm but is out of range.
Hence the maximum intensity at the hole will be for the wavelength 500 nm.
26. Consider the arrangement shown in the figure (17E-4). The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe width.
 |
| Figure for Q-26 |
ANSWER: (a) There will be a dark fringe at O if the path difference between the waves coming here from the two slits is equal to (n+½)𝜆. Let us draw a diagram as below.
 |
| Diagram for Q-26 |
Path difference Δx = ACO - ABO =2(AC-AB) =2{√(D²+d²)-D}
Hence for minimum at O,
2{√(D²+d²)-D} = (n+½)𝜆
→√(D²+d²)-D = (n+½)𝜆/2
→√(D²+d²) = D + (n+½)𝜆/2
Square both sides.
D²+d² = D² +D(n+½)𝜆 + (n+½)²𝜆²/4
Since (n+½)²𝜆²/4 is very very small in comparison to others (due to 𝜆²), we neglect it. Thus
D²+d² = D² +D(n+½)𝜆
→d² = D(n+½)𝜆
→d =√{D(n+½)𝜆}
For d to be minimum, n should be minimum i.e. n = 0.
→d = √(𝜆D/2)
(b) Let us draw a diagram.
 |
| Diagram for Q-26(b) |
At P a distance x above O, the path difference is
Δx =AC+CP-AB-BP
=(AC-AB)+(CP-BP)
Let us see first AC-AB. Considering a line parallel to ABO passing through the middle of BC.
We know that Δx =yd/D. Here y = d/2, hence
AC-AB =d²/2D
Similarly it can be shown that CP-BP ≈ (d/2-x)d/D
Since y = d/2-x
So, for maximum
d²/2D + (d/2-x)d/D =n𝜆
→d²/2D+d²/2D -xd/D =n𝜆
→d²/D -n𝜆 =xd/D
→x = d-n𝜆D/d =d -n*2d
[Since 𝜆D/2 = d², 𝜆D =2d²]
For next bright fringe n = 1
→x = d-2d =-d
So the next bright fringe will be at a distance d from O.
(c) The fringe width =2*OP =2d
27. The coherent point source S₁ and S₂ vibrating in phase emit light of wavelength 𝜆. The separation between the sources is 2𝜆. Consider a line passing through S₂ and perpendicular to the line S₁S₂. What is the smallest distance from S₂ where a minimum of intensity occurs?
ANSWER: Let there be a point P on the line S₂P such that S₂P =x. Given S₁S₂ =2𝜆. The path difference of the two waves reaching P =S₁P-S₂P
=√{x²+(2𝜆)²}-x
 |
| Diagram for Q-27 |
For a minimum,
√{x²+(2𝜆)²}-x = (n+½)𝜆
→√{x²+(2𝜆)²} = x+(n+½)𝜆
→x²+4𝜆² =x²+(n+½)²𝜆²+2x(n+½)𝜆
→2x(n+½) = 4𝜆-(n+½)²𝜆
→x = {4-(n+½)²}𝜆/2(n+½)
For n = 0, x = 15𝜆/4
for n = 1, x = {4-9/4}𝜆/3 =7𝜆/12
for n = 2, x = -9𝜆/20 {negative value not acceptable}
Hence the minimum value of x for a minimum intensity = 7𝜆/12.
28. Figure (17E-5) three equidistant slits being illuminated by a monochromatic parallel beam of light. Let BP₀-AP₀ =𝜆/3 and D>>𝜆.
(a) Show that in this case d = √(2𝜆D/3).
(b) Show that intensity at P₀ is three times the intensity due to any of the three slits individually.
 |
| Figure for Q-28 |
ANSWER: (a) Considering a horizontal reference line through the middle of AB. The distance of P₀ from this reference line =y =d/2. In this case, the path difference between BP₀ and AP₀ is
BP₀-AP₀ =y*d/D
→𝜆/3 = (d/2)*(d/D) =d²/2D
→d² = 2𝜆D/3
→d = √(2𝜆D/3)
Proved.
(b) Considering the horizontal reference line through the middle of BC, the distance of P₀ from it = d+d/2 =3d/2.
Hence the path difference between the waves coming from B and C = (3d/2)*d/D =3d²/2D =3*(2𝜆D/3)/2D =6𝜆D/6D =𝜆. So both are in the same phase.
Consider all the three waves coming at P₀. The waves from B and C are in the same phase while the wave coming from A has a phase difference
=2π(𝜆/3)/𝜆 =2π/3.
Let the intensity from each of the slits = I and corresponding maximum amplitude = r.
 |
| Diagram for Q-28 |
The resultant amplitude will be the vector sum of the individual amplitudes. Since the waves from B and C are in the same phase sum of both amplitudes = 2r. The amplitude of the wave coming from A makes an angle 2π/3 from the other two. Hence the resultant of 2r and r having an angle 2π/3 is given as
R² =(2r)²+r²+2.2r.r.cos(2π/3)
=5r²+4r²(-½) =5r²-2r² =3r²
Since the intensity is proportional to the amplitude of the wave. Hence the ratio of total intensity to one of the individual intensity = I'/I =R²/r²
→I'/I = 3r²/r² = 3
→I' = 3I
So, the intensity at P₀ is three times the intensity due to any of the three slits individually
29. In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the center of the central maximum is 0.20 W/m², what will be the intensity at a point 0.50 cm away from this center along the width of the fringes?
ANSWER: Given d = 2.0 mm = 0.002 m, D = 2.0 m, 𝜆 = 600*10⁻⁹ m =6x10⁻⁷ m.
If the intensity of each of the slits = I and the corresponding maximum amplitude of the waves = r, then amplitude at the center of the central maximum = 2r, because both the waves are in phase here. If the resultant intensity is I' then
I'/I = (2r)²/r² =4
→I = I'/4
The point in the question is at y = 0.50 cm =0.005 m. Hence the path difference between the waves
Δx = yd/D =0.005*0.002/2 m =5x10⁻⁶ m
Phase difference =2π*Δx/𝜆
=2π*5x10⁻⁶/6x10⁻⁷ =50π/3 =16π+2π/3→2π/3
The resultant amplitude here
R² = r²+r²+2.r.r.cos(2π/3) =2r²+2r²(-½) =2r²-r² =r²
→R = r
It is the amplitude corresponding to the individual intensity I. Hence the intensity at this point
I = I'/4 = 0.20/4 W/m² =0.05 W/m²
30. In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength 𝜆. Find the distance from the central point where the intensity falls to
(a) half the maximum,
(b) one-fourth of the maximum.
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'
(a) The maximum intensity at the bright fringe =4I', because of ẟ =2nπ.
The intensity half of the maximum = 2I', here
cos²(ẟ/2) =2I'/4I' =½
→cos(ẟ/2) =1/√2
→ẟ/2 =π/4
→ẟ = π/2
But Phase difference ẟ = 2π*Δx/𝜆
→Δx = ẟ𝜆/2π
So for the intensity half of the maximum, the phase difference will be π/2. And the path difference
Δx = π𝜆/4π =𝜆/4
If this point is at a distance y from the central point,
The path difference Δx =yd/D
→y =D*Δx/d =D*𝜆/4d
→y = D𝜆/4d
(b) The intensity one-fourth of the maximum =4I'/4 =I'
Hence cos²(ẟ/2) =I'/4I' =1/4
→cos(ẟ/2) = 1/2 =cos(π/3)
→ẟ/2 = π/3
→ẟ = 2π/3
Thus Δx = ẟ𝜆/2π =2π𝜆/6π =𝜆/3
Since Δx =yd/D
→𝜆/3 =yd/D
→y = D𝜆/3d
===<<<O>>>===
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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-21 TO Q-30
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- 15 - Wave Motion and Waves on a String
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CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
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CHAPTER- 11 - Gravitation
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CHAPTER- 8 - Work and Energy
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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