17: Light Waves : Neet-xii-physics

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NEET-XII-Physics

17: Light Waves

with Solutions - page 5

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Qstn #10
A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0⋅25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
Ans : Given:
Wavelengths of the source of light,
`` {\lambda }_{1}=480\times {10}^{-9}\,\mathrm{\,m\,}\,\mathrm{\,and\,}{\lambda }_{2}=600\times {10}^{-9}\,\mathrm{\,m\,}``
`` ``
Separation between the slits, `` d=0.25\,\mathrm{\,mm\,}=0.25\times {10}^{-3}\,\mathrm{\,m\,}``
Distance between screen and slit, `` D=150\,\mathrm{\,cm\,}=1.5\,\mathrm{\,m\,}``
We know that the position of the first maximum is given by
`` y=\frac{\lambda D}{d}``
So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y₂ - y₁
`` {y}_{2}-{y}_{1}=\frac{D\left({y}_{2}-{y}_{1}\right)}{d}``
`` \Rightarrow {y}_{2}-{y}_{1}=\frac{1.5}{0.25\times {10}^{-3}}\left(600\times {10}^{-9}-480\times {10}^{-9}\right)``
`` {y}_{2}-{y}_{1}=72\times {10}^{-5}\,\mathrm{\,m\,}=0.72\,\mathrm{\,mm\,}``
`` ``
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Qstn #11
White light is used in a Young’s double slit experiment. Find the minimum order of the violet fringe
λ=400 nmwhich overlaps with a red fringe
λ=700 nm.
Ans : Let the separation between the slits be d and distance between screen from the slits be D.
Suppose, the m^th bright fringe of violet light overlaps with the n^th bright fringe of red light.
Now, the position of the m^th bright fringe of violet light, y_v = `` \frac{m{\lambda }_{v}D}{d}``
Position of the n^th bright fringe of red light, y_r = `` \frac{n{\lambda }_{r}D}{d}``
For overlapping, y_{v =} y_r .
So, as per the question,
`` \frac{m\times 400\times \,\mathrm{\,D\,}}{d}=\frac{n\times 700\times \,\mathrm{\,D\,}}{d}``
`` \Rightarrow \frac{m}{n}=\frac{7}{4}``
Therefore, the `` {7}^{\,\mathrm{\,th\,}}`` bright fringe of violet light overlaps with the 4^th bright fringe of red light.
It can also be seen that the 14^th violet fringe will overlap with the 8^th red fringe.
Because, `` \frac{m}{n}=\frac{7}{4}=\frac{14}{8}``
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Qstn #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.
Ans : Given:
The refractive index of the plate is `` \mu ``.
Let the thickness of the plate be 't' to produce a change in the optical path difference of `` \frac{\lambda }{2}``.
We know that optical path difference is given by `` \left(\,\mathrm{\,\mu \,}-1\right)t``.
`` \therefore \left(\,\mathrm{\,\mu \,}-1\right)t=\frac{\lambda }{2}``
`` \Rightarrow t=\frac{\lambda }{2\left(\,\mathrm{\,\mu \,}-1\right)}``
Hence, the thickness of a plate is `` \frac{\lambda }{2\left(\,\mathrm{\,\mu \,}-1\right)}``.
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Qstn #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.
Ans : Given:
Refractive index of the plate is μ.
The thickness of the plate is t.
Wavelength of the light is λ.

#13-a
Find the change in the optical path due to introduction of the plate.
Ans : When the plate is placed in front of the slit, then the optical path difference is given by `` \left(\,\mathrm{\,\mu \,}-1\right)t``.

#13-b
What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is
λ. Neglect any absorption of light in the plate.
Ans : For zero intensity at the centre of the fringe pattern, there should be distractive interference at the centre.
So, the optical path difference should be = `` \frac{\lambda }{2}``
`` \,\mathrm{\,i\,}.\,\mathrm{\,e\,}.\left(\mu -1\right)t=\frac{\lambda }{2}``
`` \Rightarrow t=\frac{\lambda }{2\left(\mu -1\right)}``
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Qstn #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 centre if the paper is removed?
Ans : Given:
Refractive index of the paper, μ = 1.45
The thickness of the plate, `` t=0.02\,\mathrm{\,mm\,}=0.02\times {10}^{-3}\,\mathrm{\,m\,}``
Wavelength of the light, `` \lambda =620\,\mathrm{\,nm\,}=620\times {10}^{-9}\,\mathrm{\,m\,}``
`` ``
We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by `` \left(\,\mathrm{\,\mu \,}-1\right)t``.
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is
`` n=\frac{\left(\,\mathrm{\,\mu \,}-1\right)t}{\lambda }``
`` =\frac{\left(1.45-1\right)\times 0.02\times {10}^{-3}}{620\times {10}^{-9}}``
`` =14.5``
Hence, 14.5 fringes will cross through the centre if the paper is removed.
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Qstn #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 micron (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.
Ans : Given:
Refractive index of the mica sheet,μ = 1.6
Thickness of the plate, `` t=1.964\,\mathrm{\,micron\,}=1.964\times {10}^{-6}\,\mathrm{\,m\,}``
Let the wavelength of the light used = λ.
Number of fringes shifted is given by
`` n=\frac{\left(\,\mathrm{\,\mu \,}-1\right)t}{\lambda }``
So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
`` \,\mathrm{\,Shift\,}=n\times \beta ``
`` =\frac{\left(\mu -1\right)t}{\lambda }\times \frac{\lambda D}{d}``
`` =\frac{\left(\mu -1\right)t\times D}{d}...\left(\,\mathrm{\,i\,}\right)``
As per the question, when the distance between the screen and the slits is doubled,
i.e. `` D\text{'}=2D``,
fringe width, `` \beta =\frac{\lambda D\text{'}}{d}=\frac{\lambda 2D}{d}``
According to the question, fringe shift in first case = fringe width in second case.
`` \,\mathrm{\,So\,},\frac{\left(\,\mathrm{\,\mu \,}-1\right)t\times \,\mathrm{\,D\,}}{d}=\frac{\lambda 2\,\mathrm{\,D\,}}{d}``
`` \Rightarrow \lambda =\frac{\left(\mu -1\right)t}{2}``
`` =\frac{\left(1.6-1\right)\times \left(1.964\right)\times {10}^{-6}}{2}``
`` =589.2\times {10}^{-9}=589.2\,\mathrm{\,nm\,}``
Hence, the required wavelength of the monochromatic light is 589.2 nm.
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Qstn #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 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 at a distance one metre away.
Ans : Given:
The thickness of the strips = `` {t}_{1}={t}_{2}=t=0.5\,\mathrm{\,mm\,}=0.5\times {10}^{-3}\,\mathrm{\,m\,}``
Separation between the two slits,`` d=0.12\,\mathrm{\,cm\,}=12\times {10}^{-4}\,\mathrm{\,m\,}``
The refractive index of mica, μ_m = 1.58 and of polystyrene, μ_p = 1.58
Wavelength of the light, `` \,\mathrm{\,\lambda \,}=590\,\mathrm{\,nm\,}=590\times {10}^{-9}\,\mathrm{\,m\,},``
Distance between screen and slit, D = 1 m

#16-a
What would be the fringe-width?
Ans : We know that fringe width is given by
`` \beta =\frac{\lambda D}{d}``
`` \Rightarrow \beta =\frac{590\times {10}^{-9}\times 1}{12\times {10}^{-4}}``
`` =4.9\times {10}^{-4}\,\mathrm{\,m\,}``

#16-b
At what distance from the centre will the first maximum be located?
Ans : When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
`` ∆x=\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,m\,}}-1\right)t-\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,p\,}}-1\right)t``
`` =\left({\,\mathrm{\,\mu \,}}_{\,\mathrm{\,m\,}}-{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,p\,}}\right)t``
`` =\left(1.58-1.55\right)\times \left(0.5\right)\left({10}^{-3}\right)``
`` =\left(0.015\right)\times {10}^{-3}\,\mathrm{\,m\,}``
∴ Number of fringes shifted, n = `` \frac{∆x}{\lambda }``.
`` \Rightarrow n=\frac{0.015\times {10}^{-3}}{590\times {10}^{-9}}=25.43``
∴ 25 fringes and 0.43^th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43^th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
`` x=\left(0.43\right)\times 4.91\times {10}^{-4}\left(\because \beta =4.91\times {10}^{-4}\,\mathrm{\,m\,}\right)``
`` =0.021\,\mathrm{\,cm\,}``
`` ``
`` ``
On the other side,
`` x\text{'}=\left(1-0.43\right)\times 4.91\times {10}^{-4}``
`` =0.028\,\mathrm{\,cm\,}``
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Qstn #17
Two transparent slabs having equal thickness but 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?
Ans : Given:
Refractive index of the two slabs are µ₁ and µ₂.
Thickness of both the plates is t.
When both the strips are fitted, the optical path changes by
`` ∆x=\left({\,\mathrm{\,\mu \,}}_{1}-1\right)t-\left({\,\mathrm{\,\mu \,}}_{2}-1\right)t``
`` =\left({\,\mathrm{\,\mu \,}}_{1}-{\,\mathrm{\,\mu \,}}_{2}\right)t``
`` ``
For minimum at P₀, the path difference should be `` \frac{\lambda }{2}``.
`` \,\mathrm{\,i\,}.\,\mathrm{\,e\,}.∆x=\frac{\lambda }{2}``
`` \,\mathrm{\,So\,},\frac{\lambda }{2}=\left({\,\mathrm{\,\mu \,}}_{1}-{\,\mathrm{\,\mu \,}}_{2}\right)t``
`` \Rightarrow t=\frac{\lambda }{2\left({\,\mathrm{\,\mu \,}}_{1}-{\,\mathrm{\,\mu \,}}_{2}\right)}``
Therefore, minimum at point P₀ is `` \frac{\lambda }{2\left({\,\mathrm{\,\mu \,}}_{1}-{\,\mathrm{\,\mu \,}}_{2}\right)}``.
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Qstn #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.
Ans : Given:
The thickness of the thin paper, `` t=0.02\,\mathrm{\,mm\,}=0.02\times {10}^{-3}\,\mathrm{\,m\,}``
Refractive index of the paper, `` \,\mathrm{\,\mu \,}=1.45``.
Wavelength of the light, `` \lambda =600\,\mathrm{\,nm\,}=600\times {10}^{-9}\,\mathrm{\,m\,}``

#18-a
Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
Ans : Let the intensity of the source without paper = I₁
and intensity of source with paper =I₂
Let a₁ and a₂ be corresponding amplitudes.
As per the question,
`` {I}_{2}=\frac{4}{9}{I}_{1}``
We know that
`` \frac{{I}_{1}}{{I}_{2}}=\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}\left(\because I\propto {a}^{2}\right)``
`` \Rightarrow \frac{{a}_{1}}{{a}_{2}}=\frac{3}{2}``
Here, a is the amplitude.
`` \,\mathrm{\,We\,}\,\mathrm{\,know\,}\text{that}\frac{{I}_{\,\mathrm{\,max\,}}}{{I}_{\,\mathrm{\,min\,}}}=\frac{{\left({a}_{1}+{a}_{2}\right)}^{2}}{{\left({a}_{1}-{a}_{2}\right)}^{2}}.``
`` \Rightarrow \frac{{I}_{\,\mathrm{\,max\,}}}{{I}_{\,\mathrm{\,min\,}}}=\frac{{\left(3+2\right)}^{2}}{{\left(3-2\right)}^{2}}``
`` =\frac{25}{1}``
`` \Rightarrow {I}_{\,\mathrm{\,max\,}}:{I}_{\,\mathrm{\,min\,}}=25:1``

#18-b
How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
Ans : Number of fringes that will cross through the centre is given by `` n=\frac{\left(\mu -1\right)t}{\lambda }``.
`` \Rightarrow n=\frac{\left(1.45-1\right)\times 0.02\times {10}^{-3}}{600\times {10}^{-9}}``
`` =\frac{0.45\times 0.02\times {10}^{4}}{6}=15``
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