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LIGHT WAVES
EXERCISES Q-11 TO Q-20
11. White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe (𝜆 = 400 nm) which overlaps with a red fringe (𝜆 = 700 nm).
ANSWER: The fringe width is given as
w = D𝜆/d, where D - the distance of the screen from the slits, d = slit separation and 𝜆 is the wavelength of the light used.
Let n'th violet fringe overlap the p'th red fringe. The distance of the n'th violet fringe from the center
=nD*400/d nm
The distance of the p'th red fringe from the center
=pD*700/d nm.
For the violet to overlap on the red
nD*400/d = pD*700/d
→n/p = 7/4
So the first overlapping violet fringe is the 7th. Other overlappings will be in its multiple. Hence the minimum order of the violet fringe overlapping the red is 7.
11. White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe (𝜆 = 400 nm) which overlaps with a red fringe (𝜆 = 700 nm).
ANSWER: The fringe width is given as
w = D𝜆/d, where D - the distance of the screen from the slits, d = slit separation and 𝜆 is the wavelength of the light used.
Let n'th violet fringe overlap the p'th red fringe. The distance of the n'th violet fringe from the center
=nD*400/d nm
The distance of the p'th red fringe from the center
=pD*700/d nm.
For the violet to overlap on the red
nD*400/d = pD*700/d
→n/p = 7/4
So the first overlapping violet fringe is the 7th. Other overlappings will be in its multiple. Hence the minimum order of the violet fringe overlapping the red is 7.
ANSWER: The fringe width is given as
w = D𝜆/d, where D - the distance of the screen from the slits, d = slit separation and 𝜆 is the wavelength of the light used.
Let n'th violet fringe overlap the p'th red fringe. The distance of the n'th violet fringe from the center
=nD*400/d nm
The distance of the p'th red fringe from the center
=pD*700/d nm.
For the violet to overlap on the red
nD*400/d = pD*700/d
→n/p = 7/4
So the first overlapping violet fringe is the 7th. Other overlappings will be in its multiple. Hence the minimum order of the violet fringe overlapping the red is 7.
12. Find the thickness of a plate which will produce a change in optical path equal to half the wavelength 𝜆 of the light passing through it normally. The refractive index of the plate is µ.
ANSWER: The path Δx traveled in a medium of refractive index µ is equal to the path µΔx traveled in a vacuum. Hence the optical path difference = µΔx -Δx. But required path difference =𝜆/2. Hence
µΔx - Δx =𝜆/2
→Δx (µ-1)=𝜆/2
→Δx = 𝜆/{2(µ-1)}
So the light should pass through a distance Δx = 𝜆/{2(µ-1)} which is the thickness of the plate to produce an optical path difference equal to 𝜆/2.
12. Find the thickness of a plate which will produce a change in optical path equal to half the wavelength 𝜆 of the light passing through it normally. The refractive index of the plate is µ.
ANSWER: The path Δx traveled in a medium of refractive index µ is equal to the path µΔx traveled in a vacuum. Hence the optical path difference = µΔx -Δx. But required path difference =𝜆/2. Hence
µΔx - Δx =𝜆/2
→Δx (µ-1)=𝜆/2
→Δx = 𝜆/{2(µ-1)}
So the light should pass through a distance Δx = 𝜆/{2(µ-1)} which is the thickness of the plate to produce an optical path difference equal to 𝜆/2.
ANSWER: The path Δx traveled in a medium of refractive index µ is equal to the path µΔx traveled in a vacuum. Hence the optical path difference = µΔx -Δx. But required path difference =𝜆/2. Hence
µΔx - Δx =𝜆/2
→Δx (µ-1)=𝜆/2
→Δx = 𝜆/{2(µ-1)}
So the light should pass through a distance Δx = 𝜆/{2(µ-1)} which is the thickness of the plate to produce an optical path difference equal to 𝜆/2.
13. A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment.
(a) Find the change in the optical path due to the introduction of the plate.
(b) What should be the minimum thickness t which will make the intensity at the center of the fringe pattern zero?
The wavelength of the light used is 𝜆. Neglect any absorption of light in the plate.
ANSWER: (a) The optical path in thickness t of a material of refractive index µ is equivalent to µt in the vacuum. Hence the change in the optical path =µt-t =(µ-1)t
(b) For the intensity at the center of the fringe pattern to be zero, the two light waves reaching here should be completely out of phase i.e. their optical path difference should be 𝜆/2.
Hence (µ-1)t = 𝜆/2
→t = 𝜆/{2(µ-1)}
14. A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the center if the paper is removed?
13. A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment.
(a) Find the change in the optical path due to the introduction of the plate.
(b) What should be the minimum thickness t which will make the intensity at the center of the fringe pattern zero?
The wavelength of the light used is 𝜆. Neglect any absorption of light in the plate.
ANSWER: (a) The optical path in thickness t of a material of refractive index µ is equivalent to µt in the vacuum. Hence the change in the optical path =µt-t =(µ-1)t
(b) For the intensity at the center of the fringe pattern to be zero, the two light waves reaching here should be completely out of phase i.e. their optical path difference should be 𝜆/2.
Hence (µ-1)t = 𝜆/2
→t = 𝜆/{2(µ-1)}
14. A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the center if the paper is removed?
(a) Find the change in the optical path due to the introduction of the plate.
(b) What should be the minimum thickness t which will make the intensity at the center of the fringe pattern zero?
The wavelength of the light used is 𝜆. Neglect any absorption of light in the plate.
ANSWER: (a) The optical path in thickness t of a material of refractive index µ is equivalent to µt in the vacuum. Hence the change in the optical path =µt-t =(µ-1)t
(b) For the intensity at the center of the fringe pattern to be zero, the two light waves reaching here should be completely out of phase i.e. their optical path difference should be 𝜆/2.
Hence (µ-1)t = 𝜆/2
→t = 𝜆/{2(µ-1)}
14. A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the center if the paper is removed?
ANSWER: The thickness of the paper t = 0.02 mm =0.02/1000 m =2x10⁻⁵ m. The wavelength of the light 𝜆 = 620 nm = 620x10⁻⁹ m =6.2x10⁻⁷ m and the refractive index µ = 1.45, hence the optical path difference between the waves coming from two slits =(µ-1)t =(1.45-1)*2x10⁻⁵ m =9x10⁻⁶ m.
The geometrical distance of the center is equal from both the slits so no path difference, but with the introduction of a denser sheet of refractive index µ on one of the slits the optical path difference between both waves reaching the center is (µ-1)t. If this optical path difference is equal to 𝜆 then one fringe width shifts from the center.
Since for 𝜆 path difference 1 fringe shifts,
Hence for (µ-1)t path difference (µ-1)t/𝜆 fringe shifts
i.e. 9x10⁻⁶/6.2x10⁻⁷ fringe shifts
→90/6.2 fringe shifts
→≈ 14.5 fringes shift.
ANSWER: The thickness of the paper t = 0.02 mm =0.02/1000 m =2x10⁻⁵ m. The wavelength of the light 𝜆 = 620 nm = 620x10⁻⁹ m =6.2x10⁻⁷ m and the refractive index µ = 1.45, hence the optical path difference between the waves coming from two slits =(µ-1)t =(1.45-1)*2x10⁻⁵ m =9x10⁻⁶ m.
The geometrical distance of the center is equal from both the slits so no path difference, but with the introduction of a denser sheet of refractive index µ on one of the slits the optical path difference between both waves reaching the center is (µ-1)t. If this optical path difference is equal to 𝜆 then one fringe width shifts from the center.
Since for 𝜆 path difference 1 fringe shifts,
Hence for (µ-1)t path difference (µ-1)t/𝜆 fringe shifts
i.e. 9x10⁻⁶/6.2x10⁻⁷ fringe shifts
→90/6.2 fringe shifts
→≈ 14.5 fringes shift.
15. In a Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns. (1 micron = 10⁻⁶ m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
ANSWER: Let the wavelength of the monochromatic light be 𝜆. If the original screen distance be D and the slit separation =d then the distance between the successive maxima in the second case is
fringe width w =2D𝜆/d
So by this distance the fringe shifts in the first case.
Given µ = 1.6, thickness t = 1.964x10⁻⁶ m. The optical path difference at the center =(µ-1)t
=(1.6-1)*1.964x10⁻⁶ m
=1.178x10⁻⁶ m
Fringe width without the sheet in the first case =D𝜆/d
Since one fringe shifts for an optical path difference 𝜆. Here two fringes shift (2D𝜆/d) hence the optical path difference is 2𝜆. Thus
2𝜆 = 1.178x10⁻⁶ m
→𝜆 ≈ 0.590x10⁻⁶*10⁹ nm =0.590*1000 nm
→𝜆 = 590 nm
16. A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of the mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the center will the first maximum be located?
15. In a Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns. (1 micron = 10⁻⁶ m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
ANSWER: Let the wavelength of the monochromatic light be 𝜆. If the original screen distance be D and the slit separation =d then the distance between the successive maxima in the second case is
fringe width w =2D𝜆/d
So by this distance the fringe shifts in the first case.
Given µ = 1.6, thickness t = 1.964x10⁻⁶ m. The optical path difference at the center =(µ-1)t
=(1.6-1)*1.964x10⁻⁶ m
=1.178x10⁻⁶ m
Fringe width without the sheet in the first case =D𝜆/d
Since one fringe shifts for an optical path difference 𝜆. Here two fringes shift (2D𝜆/d) hence the optical path difference is 2𝜆. Thus
2𝜆 = 1.178x10⁻⁶ m
→𝜆 ≈ 0.590x10⁻⁶*10⁹ nm =0.590*1000 nm
→𝜆 = 590 nm
16. A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of the mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the center will the first maximum be located?
ANSWER: Let the wavelength of the monochromatic light be 𝜆. If the original screen distance be D and the slit separation =d then the distance between the successive maxima in the second case is
fringe width w =2D𝜆/d
So by this distance the fringe shifts in the first case.
Given µ = 1.6, thickness t = 1.964x10⁻⁶ m. The optical path difference at the center =(µ-1)t
=(1.6-1)*1.964x10⁻⁶ m
=1.178x10⁻⁶ m
Fringe width without the sheet in the first case =D𝜆/d
Since one fringe shifts for an optical path difference 𝜆. Here two fringes shift (2D𝜆/d) hence the optical path difference is 2𝜆. Thus
2𝜆 = 1.178x10⁻⁶ m
→𝜆 ≈ 0.590x10⁻⁶*10⁹ nm =0.590*1000 nm
→𝜆 = 590 nm
16. A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of the mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the center will the first maximum be located?
ANSWER: (a) The screen distance D = 1.0 m, slit separation d = 0.12 cm = 0.0012 m, 𝜆 =590 nm =5.90x10⁻⁷ m, henc ethe fringe width
w = D𝜆/d = 1*5.90x10⁻⁷/0.0012 m
→w = 4.9x10⁻⁴ m
(b) The thickness of the strips t = 0.50 mm =0.50/1000 m =5x10⁻⁴ m
The refractive indices µ = 1.58 and µ' = 1.55
The optical path difference at the center
Δx = µt -µ't =(1.58-1.55)* 5x10⁻⁴ m
→Δx = 1.5x10⁻⁵ m
Hence the number of fringe shifts from the center
= Δx/𝜆 =1.5x10⁻⁵/5.9x10⁻⁷ =25.42
Diagram for Q-16
Since 25 full fringe-widths shift from the center, it is the fractional fringe-width 0.42w on one side and (1-0.42)w =0.58w on the other side.
Hence the first maximum on one side = 0.42w
=0.42*4.9x10⁻⁴ m
=2.1x10⁻⁴*100 cm
=0.021 cm
And the first maximum on the other side =0.58w
=0.58*4.9x10⁻⁴ m
=2.8x10⁻⁴*100 cm
=0.028 cm
ANSWER: (a) The screen distance D = 1.0 m, slit separation d = 0.12 cm = 0.0012 m, 𝜆 =590 nm =5.90x10⁻⁷ m, henc ethe fringe width
w = D𝜆/d = 1*5.90x10⁻⁷/0.0012 m
→w = 4.9x10⁻⁴ m
(b) The thickness of the strips t = 0.50 mm =0.50/1000 m =5x10⁻⁴ m
The refractive indices µ = 1.58 and µ' = 1.55
The optical path difference at the center
Δx = µt -µ't =(1.58-1.55)* 5x10⁻⁴ m
→Δx = 1.5x10⁻⁵ m
Hence the number of fringe shifts from the center
= Δx/𝜆 =1.5x10⁻⁵/5.9x10⁻⁷ =25.42
 |
| Diagram for Q-16 |
Since 25 full fringe-widths shift from the center, it is the fractional fringe-width 0.42w on one side and (1-0.42)w =0.58w on the other side.
Hence the first maximum on one side = 0.42w
=0.42*4.9x10⁻⁴ m
=2.1x10⁻⁴*100 cm
=0.021 cm
And the first maximum on the other side =0.58w
=0.58*4.9x10⁻⁴ m
=2.8x10⁻⁴*100 cm
=0.028 cm
17. Two transparent slabs having equal thickness but the different refractive indices µ₁ and µ₂ are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P₀ which is equidistant from the slits?
ANSWER: The central point P₀ is equidistant from both the slits. Hence the optical path difference of the two waves coming out from the slit materials at P₀ is
=(µ₁-µ₂)t, {where t is the thickness of the slabs}
=|(µ₁-µ₂)|t, {since we need only positive value}
For a minimum at P₀ the optical path difference should be a multiple of 𝜆/2, hence for minimum t it should be 𝜆/2. Thus
|(µ₁-µ₂)|t = 𝜆/2
→t = 𝜆/{2|(µ₁-µ₂)|}
17. Two transparent slabs having equal thickness but the different refractive indices µ₁ and µ₂ are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P₀ which is equidistant from the slits?
ANSWER: The central point P₀ is equidistant from both the slits. Hence the optical path difference of the two waves coming out from the slit materials at P₀ is
=(µ₁-µ₂)t, {where t is the thickness of the slabs}
=|(µ₁-µ₂)|t, {since we need only positive value}
For a minimum at P₀ the optical path difference should be a multiple of 𝜆/2, hence for minimum t it should be 𝜆/2. Thus
|(µ₁-µ₂)|t = 𝜆/2
→t = 𝜆/{2|(µ₁-µ₂)|}
ANSWER: The central point P₀ is equidistant from both the slits. Hence the optical path difference of the two waves coming out from the slit materials at P₀ is
=(µ₁-µ₂)t, {where t is the thickness of the slabs}
=|(µ₁-µ₂)|t, {since we need only positive value}
For a minimum at P₀ the optical path difference should be a multiple of 𝜆/2, hence for minimum t it should be 𝜆/2. Thus
|(µ₁-µ₂)|t = 𝜆/2
→t = 𝜆/{2|(µ₁-µ₂)|}
18. A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it.
(a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
(b) How many fringes will cross through the center if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
ANSWER: (a) Let the intensity of light passing through the slit without the thin paper =I. Since the paper transmits only 4/9 of the light energy, hence the intensity of light transmitted by the film I₁ =4I/9.
Since the intensity of the light is proportional to the square of the amplitude, hence
I/I₁ = r²/r₁², {where r and r₁ are amplitudes of the light corresponding to the intensities I and I₁}
→r²/r₁² = I/I₁ = 9/4
→r/r₁ = 3/2
→r/3 =r₁/2 = k (say)
So r = 3k and r₁ = 2k
Maximum amplitude, R = r+r₁ = 5k,
and minimum amplitude, R' = r-r₁ = k.
If I' and I" are maximum and minimum intensities, then
I'/I" =R²/R'² =(5k)²/k² = 25
(b) The optical path difference at the center = (µ-1)t
Since for the path difference of 𝜆, 1 fringe shifts, hence in this case number of fringes passing through the center, N =(µ-1)t/𝜆
Given, µ = 1.45, thickness of paper, t = 0.02 mm =2x10⁻⁵ m, 𝜆 = 600 nm = 6x10⁻⁷ m. Hence
N = (1.45-1)*2x10⁻⁵/6x10⁻⁷ =(0.45/3)*100
→N = 0.15*100 =15
18. A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it.
(a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
(b) How many fringes will cross through the center if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
ANSWER: (a) Let the intensity of light passing through the slit without the thin paper =I. Since the paper transmits only 4/9 of the light energy, hence the intensity of light transmitted by the film I₁ =4I/9.
Since the intensity of the light is proportional to the square of the amplitude, hence
I/I₁ = r²/r₁², {where r and r₁ are amplitudes of the light corresponding to the intensities I and I₁}
→r²/r₁² = I/I₁ = 9/4
→r/r₁ = 3/2
→r/3 =r₁/2 = k (say)
So r = 3k and r₁ = 2k
Maximum amplitude, R = r+r₁ = 5k,
and minimum amplitude, R' = r-r₁ = k.
If I' and I" are maximum and minimum intensities, then
I'/I" =R²/R'² =(5k)²/k² = 25
(b) The optical path difference at the center = (µ-1)t
Since for the path difference of 𝜆, 1 fringe shifts, hence in this case number of fringes passing through the center, N =(µ-1)t/𝜆
Given, µ = 1.45, thickness of paper, t = 0.02 mm =2x10⁻⁵ m, 𝜆 = 600 nm = 6x10⁻⁷ m. Hence
N = (1.45-1)*2x10⁻⁵/6x10⁻⁷ =(0.45/3)*100
→N = 0.15*100 =15
(a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
(b) How many fringes will cross through the center if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
ANSWER: (a) Let the intensity of light passing through the slit without the thin paper =I. Since the paper transmits only 4/9 of the light energy, hence the intensity of light transmitted by the film I₁ =4I/9.
Since the intensity of the light is proportional to the square of the amplitude, hence
I/I₁ = r²/r₁², {where r and r₁ are amplitudes of the light corresponding to the intensities I and I₁}
→r²/r₁² = I/I₁ = 9/4
→r/r₁ = 3/2
→r/3 =r₁/2 = k (say)
So r = 3k and r₁ = 2k
Maximum amplitude, R = r+r₁ = 5k,
and minimum amplitude, R' = r-r₁ = k.
If I' and I" are maximum and minimum intensities, then
I'/I" =R²/R'² =(5k)²/k² = 25
(b) The optical path difference at the center = (µ-1)t
Since for the path difference of 𝜆, 1 fringe shifts, hence in this case number of fringes passing through the center, N =(µ-1)t/𝜆
Given, µ = 1.45, thickness of paper, t = 0.02 mm =2x10⁻⁵ m, 𝜆 = 600 nm = 6x10⁻⁷ m. Hence
N = (1.45-1)*2x10⁻⁵/6x10⁻⁷ =(0.45/3)*100
→N = 0.15*100 =15
19. A Young's double slit apparatus has slits separated by 0.28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light (𝜆 = 700 nm in vacuum). Find the fringe width of the pattern formed on the screen.
ANSWER: The fringe width is given as,
w =D𝜆/d. Here D = 48 cm =0.48 m, d = 0.28 mm =2.8x10⁻⁴ m. But the wavelength given is in vacuum = 700 nm while the experiment is being done in the water. Hence the wavelength in water 𝜆' = 𝜆/µ
→𝜆' = 700/1.33, {Taking µ for water 1.33}
→𝜆' = 526 nm = 5.26x10⁻⁷ m
Hence the fringe width
w =0.48*5.26x10⁻⁷/2.8x10⁻⁴ m
→w =0.90x10⁻³*1000 mm
→w = 0.90 mm
19. A Young's double slit apparatus has slits separated by 0.28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light (𝜆 = 700 nm in vacuum). Find the fringe width of the pattern formed on the screen.
ANSWER: The fringe width is given as,
w =D𝜆/d. Here D = 48 cm =0.48 m, d = 0.28 mm =2.8x10⁻⁴ m. But the wavelength given is in vacuum = 700 nm while the experiment is being done in the water. Hence the wavelength in water 𝜆' = 𝜆/µ
→𝜆' = 700/1.33, {Taking µ for water 1.33}
→𝜆' = 526 nm = 5.26x10⁻⁷ m
Hence the fringe width
w =0.48*5.26x10⁻⁷/2.8x10⁻⁴ m
→w =0.90x10⁻³*1000 mm
→w = 0.90 mm
ANSWER: The fringe width is given as,
w =D𝜆/d. Here D = 48 cm =0.48 m, d = 0.28 mm =2.8x10⁻⁴ m. But the wavelength given is in vacuum = 700 nm while the experiment is being done in the water. Hence the wavelength in water 𝜆' = 𝜆/µ
→𝜆' = 700/1.33, {Taking µ for water 1.33}
→𝜆' = 526 nm = 5.26x10⁻⁷ m
Hence the fringe width
w =0.48*5.26x10⁻⁷/2.8x10⁻⁴ m
→w =0.90x10⁻³*1000 mm
→w = 0.90 mm
20. A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle θ = sin⁻¹(𝜆/2d) with the normal to the plane of the slits, there will be a dark fringe at the center P₀ of the pattern.
ANSWER: The center P₀ is equidistant from both the slits. Hence the path difference will be due to the extra distance traveled by the wavefront to one of the slits. Since the beam is parallel, the wavefront is perpendicular to the direction of propagation. See the diagram below:-
Diagram for Q- 20
S₁ and S₂ are the slits. P₀ is the center of the pattern. S₁N is normal to the plane of slits. The incident beam makes an angle θ with this normal hence the wavefront QS₂ also makes an angle θ with the plane of the slits. QS₁ is the extra distance traveled by the wavefront in reaching the slit S₁ in comparison to S₂.
Hence the path difference = QS₁ =S₂S₂*sinθ =d.sinθ
For there be a dark fringe at P₀, this path difference should be equal to half of the wavelength of the light.
So, d.sinθ = 𝜆/2
→sinθ = 𝜆/2d
→θ = sin⁻¹(𝜆/2d)
Hence proved.
20. A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle θ = sin⁻¹(𝜆/2d) with the normal to the plane of the slits, there will be a dark fringe at the center P₀ of the pattern.
ANSWER: The center P₀ is equidistant from both the slits. Hence the path difference will be due to the extra distance traveled by the wavefront to one of the slits. Since the beam is parallel, the wavefront is perpendicular to the direction of propagation. See the diagram below:-
 |
| Diagram for Q- 20 |
S₁ and S₂ are the slits. P₀ is the center of the pattern. S₁N is normal to the plane of slits. The incident beam makes an angle θ with this normal hence the wavefront QS₂ also makes an angle θ with the plane of the slits. QS₁ is the extra distance traveled by the wavefront in reaching the slit S₁ in comparison to S₂.
Hence the path difference = QS₁ =S₂S₂*sinθ =d.sinθ
For there be a dark fringe at P₀, this path difference should be equal to half of the wavelength of the light.
So, d.sinθ = 𝜆/2
→sinθ = 𝜆/2d
→θ = sin⁻¹(𝜆/2d)
Hence proved.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
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Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
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Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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Click here for → Questions for Short Answer
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Click here for → Friction - OBJECTIVE-II
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For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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