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# HC Verma Class XII Physics Chapter 17 – Light Waves

#### Exercise : Solution of Questions on page Number : 379

White light is a composition of seven colours. These are violet, indigo, blue, green, yellow, orange and red, collectively known as VIBGYOR. Spectrum of white light consists of sever colour bands. Each band consists of some range of wavelengths or frequencies.
For orange colour : (590 nm to 620 nm)
For red colour: (620 nm to 780 nm)

So, the colour of 620 nm and 780 nm lights may be different. But the colour of 620 nm light and 621 nm light is same.

Colour of light will depend only on the frequency of light and not on the wavelength of the light. So, light will appear red to an observer under water.

The width of the central band is inversely proportional to the slit width. So, as the width of the slit is increased, the central band will become less wider and further bands will start merging in them. Hence, diffraction effects will be visible less clearly.

Light waves have the property of travelling in a straight line, unlike sound waves. When we put a cardboard between the light source and our eyes, the light waves are obstructed by the cardboard and cannot reach our eyes, which doesn’t happen when the cardboard is inserted between sound source and our ear.

To receive TV signals transmitted from Delhi in Patna directly, one has to use antennas of great height, which will cost much. On the other hand, transmission of signals with the help of satellites requires only high frequency waves and can be done easily.

Young’s double slit experiment can be performed with sound waves, as the sound waves also show interference pattern. To get a reasonable “fringe pattern”, the separation of the slits should be of the order of the wavelength of the sound waves used.
In this experiment, the bright and dark fringes can be detected by measuring the intensity of sound using a microphone or any other detector.

Interference pattern can be studied with waves of unequal intensity.
Ι=Ι1+Ι2+2[Ι1×Ι2]√cos(ϕ) ,
where ϕ=phase difference.
In case of waves of unequal intensities, the contrast will not be clear, as the minima will not be completely dark.

The interference pattern can be produced by any two coherent waves moving in the same direction. It cannot be concluded from the interference phenomenon that light is a transverse wave, as sound waves that are longitudinal in nature also interfere.

In order to get interference, the sources should be coherent, i.e. they should emit wave of the same frequency and a stable phase difference. Two candles that are placed close to each other are distinct and cannot be considered as coherent sources. Two independent sources cannot be coherent. So, two different laser sources will also not serve the purpose.

The fringe width in Young’s double slit experiment depends on the separation of the slits.
χ=λDd,
where
λ=wavelengthχ=fringe widthD=distance between slits and screend=separation between slits
On increasing d, fringe width decreases. If the separation is increased too much, the fringes will merge with each other and the fringe pattern won’t be detectable.

The violet filter will allow only violet light to pass through it. Now, if the double slit experiment is performed with the white light and violet light, the fringe pattern will not be the same as obtained by just using white light as the source. To have interference pattern, the light waves entering from the slits should be monochromatic. So, in this case, the violet light will superimpose with only violet light (of wavelength 400 nm) in such a way that the bright bands will be of violet colour and the minima will be completely dark.

(c) both a particle and a wave phenomenon

Light shows photoelectric effect and Compton effect, which depicts its particle nature. It also shows interference and diffraction, which depicts the wave nature of light.

(d) neither on elasticity nor on inertia

The speed of light in any medium depends on the refractive index of that medium, which is an intensive property. Hence, speed of light is not affected by the elasticity and inertia of the medium.

(d) electric field

Light consists of mutually perpendicular electric and magnetic fields. So, the equation of a light wave is represented by its field vector.

(d) Polarization

Reflection, interference and diffraction are the phenomena shown by both transverse waves and longitudinal waves. Polarization is the phenomenon shown only by transverse waves.

(c) its wavelength decreases but frequency remains unchanged

Frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.

Decrease in the wavelength of light entering a medium of refractive index μ is given by
λΜ=λμ,where λΜ=wavelength in medium λ=wavelength in vacuum μ=refractive index

(b) Frequency

Frequency of a light wave doesn’t change on changing the medium of propagation of light.

(a) λa>λf

An electromagnetic wave bends round the corners of an obstacle if the size of the obstacle is comparable to the wavelength of the wave. An AM wave has less frequency than an FM wave. So, an AM wave has a higher wavelength than an FM wave and it bends round the corners of a 1 m × 1m board.
λ

(d) A laser

Among the given sources, laser is the best coherent source providing monochromatic light with constant phase difference.

(d) cos−1(1/3√)
On writing the given equation in the plane equation form lx + my + nz = p,
where l2 + m2 + n2 = 1 and p>0, we get:

13√x+13√y+13√z=c3√
If θ is the angle between the normal and +X axis, then
cosθ=13√⇒θ=cos−1(13√)

(a) plane

Wave travelling from a distant source always has plane wavefront.

(a) point source

Intensity of a point source obeys the inverse square law.
Intensity of light at distance r from the point source is given by
I=S/(4πr2) ,
where S is the source strength.

(d) having a constant phase difference
For light waves emitted by two sources of light to remain coherent, the initial phase difference between waves should remain constant in time. If the phase difference changes continuously or randomly with time, then the sources are incoherent.

(d) interference of light

Interference effect is produced by a thin film ( coating of a thin layer of a translucent material on a medium of different refractive index which allows light to pass through it))In the present case, oil floating on water forms a thin film on the surface of water, leading to the display of beautiful colours in daylight because of the interference of sunlight.

#### Exercise : Solution of Questions on page Number : 380

Given:
Range of wave length is
400 nm<λ<700 nm
We know that frequency is given by f=cλ,
where c=speed of light=3×108 m/s
f is the frequency
λ is the wavelength
We can write wavelength as:
1700 nm<1λ<1400 nm⇒17×10-7 m<1λ<14×10-7 m3×1087×10-7 Hz<cλ<3×1084×10-7 Hz⇒4.3×1014 Hz<cλ<7.5×1014 Hz⇒4.3×1014 Hz<f<7.5×1014 Hz

Hence, frequency of the range of light that is visible to an average human being is 4.3×1014 Hz to7.5×1014 Hz.

Given:
Wavelength of sodium light in air, λa=589 nm=589×10-9 m
Refractive index of water, μw= 1⋅33
We know that f=cλ,
where c=speed of light=3×108 m/s
f = frequency
λ = wavelength
(a) Frequency in air, fair=cλa
fair=3×108589×10-9 =5.09×1014 Hz

(b)
Let wavelength of sodium light in water be λw.
We know that
μaμw=λωλa,
where μa is the refractive index of air which is equal to 1 and
λw is the wavelength of sodium light in water.
⇒11.33=λω589×10-9⇒ λω=443 nm

(c) Frequency of light does not change when light travels from one medium to another.
∴  fω=fa =5.09×1014 Hz

(d) Let the speed of sodium light in water be νω
and speed in air, va = c.
Using μaμω=νωνa, we get:
νω=μacμω =3×1081.33=2.25×108 m/s

Given:
Refractive index of fused quartz for light of wavelength 400 nm is 1.472.
And refractive index of fused quartz for light of wavelength 760 nm is 1.452.
We known that refractive index of a material is given by
μ = Speed of light in vacuumSpeed of light in the material=cv
Let speed of light for wavelength 400 nm in quartz be v400.
So,
1.472=3×108v400⇒v400=2.04×108 m/s
Let speed of light of wavelength 760 nm in quartz be v760.
Again, 1.4521=3×108ν760
⇒v760=2.07×108 m/s

Given:
Speed of yellow light in liquid (vL)= 2⋅4 × 108 m s−1
And speed of yellow light in air speed = va
Let μL be the refractive index of the liquid
Using, μL=speed of light in vaccumvelocity of light in the given medium=cvL
μL=3×1082.4×108=1.25

Hence, the required refractive index is 1.25.

Given:
Separation between two narrow slits, d = 1 cm = 10−2 m
Wavelength of the light, λ=5×10-7 m
Distance of the screen, D=1 m
(a)
We know that separation between two consecutive maxima = fringe width (β).
That is, β=λDd …(i)
=5×10-7×110-2 m =5×10-5 m=0.05 mm

(b)
Separation between two consecutive maxima = fringe width
∴  β=1 mm=10-3 m
Let the separation between the sources be ‘d’
Using equation (i), we get:
d’=5×10-7×110-3 ⇒d’=5×10-4 m=0.50 mm

(c) 9 : 4

Ratio of maximum intensity and minimum intensity is given by
ImaxImin=I1 +I22I1-I22=251⇒I1=3 and I2=2⇒I1=9 and I2=4
Then,
I1I2=94

(b) I0/​4

Total intensity coming from the source is I0 which is present at the central maxima. In case of two slits, the intensity is getting distributed between the two slits and for a single slit, the amplitude of light coming from the slit is reduced to half which leads to 1/4th of intensity.

(c) remain same

On the introduction of a transparent sheet in front of one of the slits, the fringe pattern will shift slightly but the width will remain the same.

(a) the fringe width will decrease

As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,
Here, λΜ=λ/ηλΜ=wavelength in mediumλ=wavelength in vacuumη=refractive index of medium
Hence, fringe width decreases when Young’s double slit experiment is performed under water.

(a) in vacuum
(c) in a material medium

Light is an electromagnetic wave that can travel through vacuum or any optical medium.

(b) Speed of light in water is smaller than its speed in vacuum.
(c) Light shows interference.

Snell’s Law, which states that the speed of light reduces on moving from a rarer to a denser medium, can be concluded from Huygens’ wave theory and interference of light waves is based on the wave properties of light.

(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular

Light is an electromagnetic wave that propagates through its electric and magnetic field vectors, which are mutually perpendicular to each other, as well as to the direction of propagation of light. The average value of both the fields is zero.

(c) find the new position of a wavefront
(d) explain Snell’s Law

Huygen’s wave theory explains the origin of points for the new wavefront proceeding successively. It also explains the variation in speed of light on moving from one medium to another, i.e. it proves Snell’s Law.

(c) vA = vB = vC
(d) νB=12vA+vC

Since the speed of light is a universal constant, vA = vB = vC = 3×108 m/s.
νB=12vA+vC. This expression also implies that vA = vB = vC.
1
νs=12

(a) vA > vB > vC
(d) vB = (vA + vC )/2

In any other medium, the speed of light is given by v=c/η, where η is the refractive index of the medium and according to Doppler effect, for an observer moving towards the source ,speed of light appears to be more than the other two cases. On the other hand, it will be least when the observer is moving away from the source.

(a) x = c

​The wave is travelling along the X-axis. So, it’ll have planar wavefront perpendicular to the X-axis.

(b) consecutive fringes will come closer

Fringe width, β=λD/d.
Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(d) The fringe next to the central will be violet.

The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.

(a) (i) and (ii)
(d) (iii) and (iv).

The waves are travelling with the same frequencies and varying by constant phase difference.

<h4″>Exercise : Solution of Questions on page Number : 381

Given:
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10−3 m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10−3 m
Let the wavelength of the light used in experiment be λ.
We know that
β=λDd
10-3 m=2.5×λ10-3⇒λ=12.5 10-6 m =4×10-7 m=400 nm

Hence, the wavelength of light used for the experiment is 400 nm.

Given:
Separation between the two slits, d=1 mm=10-3 m
Wavelength of the light used, λ=5.0×10-7 m
Distance between screen and slit, D=1 m
(a) The distance of the centre of the first minimum from the centre of the central maximum, x = width of central maxima2
That is, x=β2=λD2d …(i)
=5×10-7×12×10-3 =2.5×10-4 m=0.25 mm
(b) From equation (i),
fringe width, β=2×x=0.50 mm
So, number of bright fringes formed in one centimetre (10 mm) = 100.50=20.

Given:
Separation between two narrow slits, d=0.8 mm=0.8×10-3 m
Wavelength of the yellow light, λ=589 nm=589×10-9 m
Distance between screen and slit, D=2.0 m
Separation between the adjacent bright bands = width of one dark fringe
That is, β=λDd …(i)
⇒β=589×10-9×20.8×10-3 =1.47×10-3 m =1.47 mm.

Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.

Given:
Wavelength of the blue-green light, λ=500×10-9 m
Separation between two slits, d=2×10-3 m,
Let angular separation between the consecutive bright fringes be θ.
Using θ= βD=λDdD=λd, we get: θ=500×10-92×10-3 =250×10-6 =25×10-5 radian or 0.014°
Hence, the angular separation between the consecutive bright fringes is 0.014 degree.

Given:
Wavelengths of the source of light,
λ1=480×10-9 m and λ2=600×10-9 m
Separation between the slits, d=0.25 mm=0.25×10-3 m
Distance between screen and slit, D=150 cm=1.5 m
We know that the position of the first maximum is given by
y=λDd
So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y2 − y1
y2-y1=Dy2-y1d

⇒y2-y1=1.50.25×10-3600×10-9-480×10-9 y2-y1=72×10-5 m =0.72 mm

Let the separation between the slits be d and distance between screen from the slits be D.
Suppose, the mth bright fringe of violet light overlaps with the nth bright fringe of red light.
Now, the position of the mth bright fringe of violet light, yv = mλvDd
Position of the nth bright fringe of red light, yr = nλrDd
For overlapping, yv = yr .
So, as per the question,
m×400×Dd=n×700×Dd⇒mn=74
Therefore, the 7th bright fringe of violet light overlaps with the 4th bright fringe of red light.
It can also be seen that the 14th violet fringe will overlap with the 8th red fringe.
Because, mn=74=148

Given:
The refractive index of the plate is μ.
Let the thickness of the plate be ‘t’ to produce a change in the optical path difference of λ2.
We know that optical path difference is given by μ-1t.
∴ μ-1t=λ2⇒t=λ2μ-1
Hence, the thickness of a plate is λ2μ-1.

Given:
Refractive index of the plate is μ.
The thickness of the plate is t.
Wavelength of the light is λ.
(a)
When the plate is placed in front of the slit, then the optical path difference is given by μ-1t.

(b) 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 = λ2
i.e. μ-1 t=λ2⇒t=λ2 μ-1

Given:
Refractive index of the paper, μ = 1.45
The thickness of the plate, t=0.02 mm=0.02×10-3 m
Wavelength of the light, λ=620 nm=620×10-9 m
We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by μ-1t.
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is
n=μ-1tλ =1.45-1×0.02×10-3620×10-9 =14.5

Hence, 14.5 fringes will cross through the centre if the paper is removed.

Given:
Refractive index of the mica sheet,μ = 1.6
Thickness of the plate, t=1.964 micron=1.964×10-6 m
Let the wavelength of the light used = λ.
Number of fringes shift Answer : 41 d is given by
n=μ-1tλ
So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
Shift =n × β=μ-1tλ×λDd=μ-1t×Dd …(i)
As per the question, when the distance between the screen and the slits is doubled,
i.e. D’=2D,
fringe width, β=λD’d=λ2Dd
According to the question, fringe shift in first case = fringe width in second case.
So, μ-1t×Dd=λ2Dd⇒λ=μ-1 t2 =1.6-1×1.964×10-62 =589.2×10-9=589.2 nm

Hence, the required wavelength of the monochromatic light is 589.2 nm.

Given:
The thickness of the strips = t1=t2=t=0.5 mm=0.5×10-3 m
Separation between the two slits, d=0.12 cm=12×10-4 m
The refractive index of mica, μm = 1.58 and of polystyrene, μp = 1.58
Wavelength of the light, λ=590 nm=590×10-9 m,
Distance between screen and slit, D = 1 m

(a)
We know that fringe width is given by
β=λDd
⇒β =590×10-9×112×10-4 =4.9×10-4 m

(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
∆x=μm-1 t- μp-1 t =μm-μp t =1.58-1.55×0.5 10-3 =0.015×10-3 m
∴  Number of fringes shifted, n = ∆xλ.
⇒ n=0.015×10-3590×10-9=25.43
∴  25 fringes and 0.43 th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
x=0.43×4.91×10-4 ∵β=4.91×10-4 m =0.021 cm
On the other side,
x’=1-0.43×4.91×10-4 =0.028 cm

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=μ1-1 t- μ2-1 t =μ1-μ2 t
For minimum at P0, the path difference should be λ2.
i.e.  ∆x=λ2
So,λ2=μ1-μ2t⇒t=λ2μ1-μ2
Therefore, minimum at point P0 is λ2μ1-μ2.

Given:
The thickness of the thin paper, t=0.02 mm=0.02×10-3 m
Refractive index of the paper, μ=1.45.
Wavelength of the light, λ=600 nm=600×10-9 m
(a)
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,
I2=49I1
We know that
I1I2=a12a22
∵ I∝a2⇒a1a2=32
Here, a is the amplitude.
We know that ImaxImin=a1+a22a1-a22.⇒ ImaxImin=3+223-22 =251⇒Imax:Imin=25 : 1

(b)
Number of fringes that will cross through the centre is given by n=μ-1tλ.
⇒ n=1.45-1×0.02×10-3600×10-9 =0.45×0.02×1046=15

Given:
Separation between two slits, d=0.28 mm=0.28×10-3 m
Distance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light, λa=700 nm in vaccum =700 ×10-9 m
Let the wavelength of red light in water = λω
We known that refractive index of water (μw =4/3),
μw = Speed of light in vacuumSpeed of light in the water
So, μw=vavω=λaλω⇒43=λaλω⇒λω= 3λa4=3×7004=525 nm
So, the fringe width of the pattern is given by
β=λωDd =525×10-9×0.480.28×10-3 =9×10-4=0.90 mm

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.

Let the two slits are S1 and S2 with separation d as shown in figure. The wave fronts reaching P0 from S1 and S2 will have a path difference of S1X = ∆x.
In ∆S1S2X, sinθ=S1XS1S2=∆xd
⇒∆x=dsinθ
Using θ=sin-1 λ2d, we get,
⇒∆x=d×λ2d=λ2
Hence, there will be dark fringe at point P0 as the path difference is an odd multiple of λ2.

(a) The phase of a light wave reflecting from a surface differs by ‘π’ from the light directly coming from the source. Thus, the wave fronts reaching just above the mirror directly from the source and after reflecting from the mirror have a phase difference of π, which is the condition of distractive interference. So, the intensity at a point just above the mirror is zero.

(b) Here, separation between two slits is 2d.
Wavelength of the light is λ.
Distance of the screen from the slit is D.
Consider that the bright fringe is formed at position y. Then,
path difference, ∆x=y×2dD=nλ.
After reflection from the mirror, path difference between two waves is λ2.
⇒ y×2dD=λ2+nλ For first order, put n=0.⇒y=λD4d

#### Exercise : Solution of Questions on page Number : 382

Given:
Separation between two slits, d’=2d=2 mm=2×10-3 m (as d = 1 mm)
Wavelength of the light used, λ=700 nm=700×10-3 m
Distance between the screen and slit (D) = 1.0 m
It is a case of Lloyd’s mirror experiment.
Fringe width, β=λDd’ =700×10-9×12×10-3 =0.35×10-3 m=0.35 mm

Hence, the width of the fringe is 0.35 mm.

Given:
The mirror reflects 64% of the energy or intensity of light.
Let intensity of source = I1.
And intensity of light after reflection from the mirror = I2.
Let a1 and a2 be corresponding amplitudes of intensity I1 and I2.
According to the question,
I2=I1×64100⇒I2I1=64100=1625And I2I1=a22a12⇒a2a1=45
We know that Imax Imin=a1+a22a1-a22 =5+425-42 Imax : Imin=81 : 1

Hence, the required ratio is 81 : 1.

Given:
Separation between the two slits = d
Wavelength of the coherent light =λ
Distance between the slit and mirror is D1.
Distance between the slit and screen is D2.
Therefore,
apparent distance of the screen from the slits,
D=2D1+D2Fringe width, β=λDd=2D1+D2 λd

Hence, the required fringe width is 2D1+D2 λd.

Given: Separation between two slits,
d=0.5 mm=0.5×10-3 m
Wavelength of the light, λ=400 nm to 700 nm
Distance of the screen from the slit, D=50 cm=0.5 m,
Position of hole on the screen, yn=1 mm=1×10-3 m

(a) The wavelength(s) will be absent in the light coming from the hole, which will form a dark fringe at the position of hole.
yn=2n+1λn2Dd , where n = 0, 1, 2, …

⇒λn=22n+1 yndD =22n+1×10-3×0.05×10-30.5 =22n+1×10-6 m =22n+1×103 nm

For n=1,λ1=23×1000 =667 nmFor n=2,λ2=25×1000=400 nm
Thus, the light waves of wavelength 400 nm and 667 nm will be absent from the light coming from the hole.

(b) The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.
So, yn=nλnDd⇒λn=yndnDFor n=1, λ1=yndD =10-3×0.5×10-30.5 =10-6 m=1000 nm.

But 1000 nm does not fall in the range 400 nm − 700 nm.

Again, for n=2,λ2=ynd2D=500 nm
So, the light of wavelength 500 nm will have strong intensity. From the figure, AB=BO and AC=CO.
Path difference of the wave front reaching O,
∆x=AB+BO-AC+CO =2 AB-AC =2 D2+d2-D
For dark fringe to be formed at O, path difference should be an odd multiple of λ2.
So,∆x=2n+1 λ2⇒ 2D2+d2-D=2n+1 λ2⇒D2+d2=D+2n+1 λ4⇒D2+d2=D2+2n+12λ216+2n+1 λD2
Neglecting, 2n+12λ216, as it is very small,
we get :
d=2n+1 λD2For minimum d, putting, n=0dmin=λD2, we get:
Thus, for dmin=λD2 there is a dark fringe at O. Let P be the point of minimum intensity.

For minimum intensity at point P,

S1P-S2P=x=2n+1 λ2

Thus, we get:

Z2+2λ2-Z=2n+1 λ2⇒Z2+4λ2=Z2 2n+12 λ24+2Z 2n+1 λ2⇒Z=4λ2-2n+1 2λ2/42n+1 λ =16λ2-2n+14 2n+1 λ … (i)

When n=0, Z=15λ4 n=-1, Z=-15λ4 n=1, Z=7λ12 n=2 , Z=-9λ20

Thus, Z=7λ 12 is the smallest distance for which there will be minimum intensity.

(a) Given:
Wavelength of light = λ
Path difference of wave fronts reaching from A and B is given by
∆xB=BP0-AP0=λ3⇒D2+d2-D=λ3⇒D2+d2= D2+λ29 +2λD3

We will neglect the term λ29, as it has a very small value.
∴ d=2λD3

(b) To calculating the intensity at P0, consider the interference of light waves coming from all the three slits.
Path difference of the wave fronts reaching from A and C is given by
CP0-AP0=D2+2d2-D =D2+8λD3-D Using the value of d from part a =D1+8λ3D12-D Expanding the value using binomial theorem and neglecting the higher order terms, we get : D1+12×8λ3D+…-D
CP0-AP0=4λ3
So, the corresponding phase difference between the wave fronts from A and C is given by
ϕc=2π∆xCλ=2π×4λ3λ⇒ϕc=8π3 or 2π+2π3⇒ϕc=2π3 …(1)
Again, ϕB=2π ∆xBλ⇒ϕB=2πλ3λ=2π3 …(2)

So, it can be said that light from B and C are in the same phase, as they have the same phase difference with respect to A.

Amplitude of wave reaching P0 is given by
A=2a2+a2+2a×acos2π3 =4a2+a2+2a23
∴ lpo=K 3 r2=3 Kr2=3l
Here, I is the intensity due to the individual slits and Ipo is the total intensity at P0.
Thus, the resulting amplitude is three times the intensity due to the individual slits. Given:
Separation between the slits, d=2 mm=2×10-3 m
Wavelength of the light, λ=600 nm=6×10-7 m
Distance of the screen from the slits, D = 2⋅0 m
Imax=0.20 W/m2For the point at a position y=0.5 cm=0.5×10-2 m,path difference,∆x=ydD.⇒∆x=0.5×10-2×2×10-32 =5×10-6 m

So, the corresponding phase difference is given by
∆ϕ=2π∆xλ=2π×5×10-66×10-7 =50π3=16π+2π3or ∆ϕ=2π3
So, the amplitude of the resulting wave at point y = 0.5 cm is given by
A=a2+a2+2a2 cos 2π3 =a2+a2-a2 =a
Similarly, the amplitude of the resulting wave at the centre is 2a.
Let the intensity of the resulting wave at point y = 0.5 cm be I.
Since IImax=A22a2, we have:I0.2=A24a2=a24a2⇒I=0.24=0.05 W/m2
Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m2.

Given:
Separation between the two slits = d
Wavelength of the light = λ
Distance of the screen = D
(a) When the intensity is half the maximum:
Let Imax be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So, I=a2+a2+2a2cosϕ
Here, ϕ is the phase difference in the waves coming from the two slits.
So, I=4a2cos2ϕ2⇒IImax=12⇒4a2cos2ϕ24a2=12⇒cos2ϕ2=12⇒cosϕ2=12⇒ϕ2=π4⇒ϕ=π2
Corresponding path difference, ∆x=λ4⇒y=∆xDd=λD4d

(b) When the intensity is one-fourth of the maximum:

IImax=14⇒4a2cos2ϕ2=14⇒cos2 ϕ2=14⇒cosϕ2=12⇒ϕ2=π3So, corrosponding path difference, ∆x=λ3and position, y=∆xDd=λD3d .

Given:
Separation between the two slits, d=1 mm=10-3 m
Wavelength of the light, λ=500 nm=5×10-7 m
Distance of the screen, D=1 m
Let Imax be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So, I=a2+a2+2a2cosϕ
Here, ϕ is the phase difference in the waves coming from the two slits.
So, I=4a2cos2ϕ2
⇒ IImax=12⇒4a2cos2ϕ24a2=12⇒cos2ϕ2=12⇒cosϕ2=12⇒ϕ2=π4⇒ϕ=π2
Corrosponding path difference, ∆x=14⇒y=∆xDd=λD4d
⇒ y=5×10-7×14×10-3 =1.25×10-4 m

∴  The required minimum distance from the central maximum is 1.25×10-4 m.

Given:
Separation between two slits = d
Wavelength of the light = λ
Distance of the screen = D
Let Imax be the maximum intensity and I be half the maximum intensity at a point at a distance y from the central point.
So, I=a2+a2+2a2cosϕ
Here, ϕ is the phase difference in the waves coming from the two slits.
So, I=4a2cos2ϕ2
⇒IImax=12⇒4a2cos2ϕ24a2=12⇒cos2ϕ2=12⇒cosϕ2=12⇒ϕ2=π4⇒ϕ=π2Corrosponding path difference, ∆x=14⇒y=∆xDd=λD4d
The line-width of a bright fringe is defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum.
So, line-width = 2y
=2Dλ4d=Dλ2d

Thus, the required line width of the bright fringe is Dλ2d.

#### Exercise : Solution of Questions on page Number : 383

Given:
Separation between the two slits = d
Wavelength of the light =λ
Distance of the screen = D
The fringe width (β) is given by β=λDd.
At S3, the path difference is zero. So, the maximum intensity occurs at amplitude = 2a.
(a) When z=Dλ2d:
The first minima occurs at S4, as shown in figure (a).
With amplitude = 0 on screen ∑2, we get:
lmaxlmin=2a+022a-02=1 (b) When z=Dλd:
The first maxima occurs at S4, as shown in the figure. With amplitude = 2a on screen ∑2,
we get : lmaxlmin=2a+2a22a-2a2=∞

(c) When z=Dλ4d: The slit S4 falls at the mid-point of the central maxima and the first minima, as shown in the figure.

Intensity=lmax2⇒ Amplitude=2a
∴ lmaxlmin=2a+2a22a-2a2=34

Given:
Fours slits S1, S2, S3 and S4.
The separation between slits S3 and S4 can be changed.
Point P is the common perpendicular bisector of S1S2 and S3S4. (a) For z = λDd :
The position of the slits from the central point of the first screen is given by
y=OS3=OS4=z2=λD2d
The corresponding path difference in wave fronts reaching S3 is given by
∆x=ydD=λD2d×dD=λ2
Similarly at S4, path difference, ∆x=ydD=λD2d×dD=λ2
i.e. dark fringes are formed at S3 and S4.
So, the intensity of light at S3 and S4 is zero. Hence, the intensity at P is also zero.

(b) For z=3λD2d
The position of the slits from the central point of the first screen is given by
y=OS3=OS4=z2=3λD4d
The corresponding path difference in wave fronts reaching S3 is given by
∆x=ydD=3λD4d×dD=3λ4
Similarly at S4, path difference, ∆x=ydD=3λD4d×dD=3λ4
Hence, the intensity at P is I.

(c) Similarly, for z=2Dλd,
the intensity at P is 2I.

Given:
Thickness of soap film, d =0.0011 mm = 0.0011 × 10−3 m
Wavelength of light used, λ=580 nm=580×10-9 m
Let the index of refraction of the soap solution be μ.
The condition of minimum reflection of light is 2μd = nλ,
where n is an interger = 1 , 2 , 3 …
⇒μ=nλ2d=2nλ4d =580×10-9×2n4×11×10-7 =5.82n44=0.1322n
As per the question, μ has a value between 1.2 and 1.5. So,
n=5So, μ=0.132×10=1.32

Therefore, the index of refraction of the soap solution is 1.32.

Given:
Wavelength of light used, λ=560×10-9 m
Refractive index of the oil film, μ=1.4
Let the thickness of the film for strong reflection be t.
The condition for strong reflection is
2μt=2n+1λ2⇒t=2n+1λ4μ
where n is an integer.
For minimum thickness, putting n = 0, we get:
t=λ4μ =560×10-94×1.4 =10-7m=100 nm
Therefore, the minimum thickness of the oil film so that it strongly reflects the light is 100 nm.

Given,
Wavelength of light used, λ=400×10-9 m to 700×10-9 nm
Refractive index of water, μ=1.33
The thickness of film, t=10-4 cm=10-6 m
The condition for strong transmission: 2μt=nλ,
where n is an integer.
⇒ λ=2μtn
⇒ λ=2×1.33×10-6n =2660×10-9n m
Putting n = 4, we get, λ1 = 665 nm.
Putting n = 5, we get, λ2 = 532 nm.
Putting n = 6, we get, λ3 = 443 nm.
Therefore, the wavelength (in visible region) which are strongly transmitted by the film are 665 nm, 532nm and 443 nm.

Given:
Wavelength of light used, λ=400×10-9 to 750×10-9 m
Refractive index of oil, μoil, is 1.25 and that of glass, μg, is 1.50.
The thickness of the oil film, d=1×10-4 cm=10-6m,
The condition for the wavelengths which can be completely transmitted through the oil film is given by
λ=2μdn+12 =2×10-6×1.25×22n+1 =5×10-6 2n+1 m⇒ λ=50002n+1 nm
Where n is an integer.
For wavelength to be in visible region i.e (400 nm to 750 nm)
When n = 3, we get,
λ=50002×3+1 =50007=714.3 nm
When, n = 4, we get,
λ=50002×4+1 =50009=555.6 nm
When, n = 5, we get,
λ=50002×5+1 =500011=454.5 nm
Thus the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455 nm.

Given:
Width of the slit, b = 5.0 cm
First diffraction minimum is formed at θ = 30°.
For the diffraction minima, we have:
bsinθ = nλ
For the first minima, we put n = 1.
5×sin30°=1×λ⇒λ=52=2.5 cm
Therefore, the wavelength of the microwaves is 2.5 cm.

Given:
Wavelength of the light used, λ=560 nm=560×10-9 m
Diameter of the pinhole, d = 0.20 mm = 2 × 10−4 m
Distance of the wall, D = 2m
We know that the radius of the central bright spot is given by
R=1.22λDd =1.22×560×10-9×22×10-4 =6.832×10-3 m or=0.683 cm

Hence, the diameter 2R of the central bright spot on the wall is 1.37 cm.