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cr
ac
ki it
je
e. in
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Table of Contents
Interference _________________________________________________________________ 2 Analytical Interference______________________________________________________________ 3 Condition for minimum Intensity _____________________________________________________ 4 Condition for Constrictive Interference_________________________________________________ 5 Condition for Destrictive Interference _________________________________________________ 5 Summary of Interference ____________________________________________________________ 8
Learning Points ______________________________________________________________ 14
e. in
Concept of Optical Equivalence _________________________________________________ 15 Optical Instruments __________________________________________________________ 22 Microscope _________________________________________________________________ 22
je
Uses____________________________________________________________________________ 25
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Compound Microscope ____________________________________________________________ 25
Telescope __________________________________________________________________ 28 Terrestrial Telescope ______________________________________________________________ 28
ac
Galilean Telescope ________________________________________________________________ 30 Astronomical Telescope ____________________________________________________________ 32
cr
Geometry of Fringes _______________________________________________________________ 35 Intensity Distribution ______________________________________________________________ 36 Black Line _______________________________________________________________________ 36
Coherent Sources ____________________________________________________________ 37 Young’s Double Slit Experiment _________________________________________________ 41 Fringe width () __________________________________________________________________ 43 Angular Fringe width () ___________________________________________________________ 44 Geometry of Fringes _______________________________________________________________ 44 Intensity Distribution ______________________________________________________________ 44
Examples ___________________________________________________________________ 47
1
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Interference The phenomenon of redistribution of energy when two waves travelling in same direction or rarely same direction meet each other is called interference.
Interface
Principle of Superposition of waves
e. in
Law of Conservation of energy
je
Redistribution of energy
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Constructive Interference – waves help each other.
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ac
Destructive Interference – waves oppose each other.
Interference
Constructive
Destructive
2
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Analytical Interference 1 = a1 sin (wt-kx) 2 = a2 sin (wt-kx+) W = 2f 1 = a1 sin wt 2 = a2 sin (wt+) Phase difference = 2 - 1 =
= 1 + 2 = a1 sin wt + a2 sin (wt+)
= a sin (wt+)
ki it
je
= a1 sin wt + a2 sin wt cos + a2 sin cos wt)
e. in
From principle of superposition
A’ resultant Amplitude = a12 a22 2a1a2 cos
cr
ac
a2 sin tan1 a1 a2 cos
When cos = 1 Amax = a1 + a2
When cos = – 1 Amax = a1 – a2 Let a1 = a2 = a Amax = 2a Amin = 0
3
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Condition for Constructive interference cos = 1
= phase difference = even multiple of
Condition for destructive interference cos = - 1 = phase difference = odd multiple of
e. in
Intensity Relation I a2
ki it
je
I = I1 + I2 +
cos = 1 +
)2
cr
Imin = (
ac
Condition for maximum Intensity
Condition for minimum Intensity cos = - 1 Imax = (
-
)2
Let a1 = a2 I1 = I2 Imax = 2I Imin = 0 Path difference = 2 Phase difference 4
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Δx = Path difference Δ = Phase difference
x
2
Condition for Constrictive Interference Δx = Path difference = n
e. in
= even multiple of /2
je
Condition for Destrictive Interference Δx = odd multiple of /2
ki it
Let I1 = I2 = I0
Let Imax = 4I0
ac
IP = 4 I0 cos2(/2)
cr
IP = 4 Imax cos2(/2)
5
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e. in
Example: A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in fig. Another identical glass plate is kept close to the first one–and parallel to it. Each glass plate reflects 25 percent of the light incident on it and transmits the remaining. Find the ratio of the minimum and maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.
I1 = 0.25 I = I/4 I2 = (0.75)(0.25)(0.75I) 9I
I
I1 I2 I1
2
2
2
49 Ans. 1
cr
Imax Imin
ac
64
ki it
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Solution
6
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λ path difference 2 phase difference Δ x = path difference Δ = path difference
Δx=
Δ
λ
= (2n) = (even multiple)
λ
e. in
Condition for Constructive Interference: Δ = (even multiple) Δx=nλ = (integral multiple) λ Condition of Destructive Interference: λ
ki it
Let I1 = I2 = I0
je
Δ x = (odd multiple)
ac
IP = 4 I0 cos2
= phase difference between the arriving waves.
∴
cr
Imax = 4 I0
IP = 4 Imax cos2
7
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Summary of Interference If amplitudes of waves arriving at point P on the screen are different then resultant intensity is given by I=
Imax =
Also,
I2
2
I1
,
Imin =
I1
I2
2
,
I1 I1
I2 I2
2
= =
AA AA = rr 11
Im ax I m in
(ii)
I1 I 2 2 I1 I 2 cos
where
when
2
1
2
1
2
r=
A1 A2
=
when
cos = 1 cos = – 1
2
e. in
(i)
I1 I2
ac
ki it
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(iii) The phenomenon of interference is based on conservation of energy. There is no destruction of energy in the interference phenomenon. The energy which apparently disappears at the minima has actually been transferred to the maxima where the intensity is greater than that produced by the two beams acting separately. 2
0 2
cr
Iav =
I d d
2
= 21
(I
1
I 2 2 I 1 I 2 cos ) d = I1 + I2
0
0
2
cos d = 0 0
As the average value of intensity is equal to the sum of individual intensities, therefore the energy is not destroyed but merely redistributed in the interference pattern.
8
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If whole apparatus is immersed in liquid of refractive index then,
(ii)
D d
i.e., fringes width decreases
ki it
=
je
(i)
e. in
(iv) All maxima are equally spaced and equally bright. This is true for minima as well. Also interference maxima and minima are alternate. The intensity distribution in interference pattern is shown in fig.
Sometimes in numerical problems, angular fringe width () is given which is defined as angular separation between two consecutive maxima or minima. D d
ac
=
cr
In medium, other than air or vacuum, =
(iii)
d
yd valid when angular position of maxima or minima is less than is . 6 D However x d sin is valid for larger values of provided d 1. Equation (1) and (4) also show that for
e. in
magnification since the factor
je
(i) the focal length ‘ƒO’ of the objective should be large, and
ki it
(ii)the focal length ‘ƒe’ of the eyepiece should be small.
cr
ac
The objective of an astronomical telescope should have a large aperture so as to allow a large number of rays to fall on it. In that case, the intensity of the image is large. This becomes even more essential in the case of heavenly bodies which are situated at large distances. A powerful astronomical telescope placed at Yerke’s observatory at Lake Geneva has an objective aperture about one metre and of focal length about 18 metre.
Geometry of Fringes For any fringe Δx = constant S1P – S2P = constant
The locus of any point which moves in such a way so that difference of its distance from the two paired point is constant is known as Hyperbola. 35
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Intensity Distribution IS1 = IS2 = I (Say) IP 4Icos2( / 2)
Where = Phase difference
Hyperbola approaching straight line for large values of x. R > V
e. in
R > V = Wavelength of light in air ’ = Wavelength of light in medium
je
ki it
=
cr
ac
Black Line This is defined as points where minima are due to two wavelengths coincide.
36
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Example: Two sources S1 and S2 give out waves of same frequency separated by 3 as shown in fig. Find the number of maxima and minima recorded on straight screen passing through S1 and perpendicular line joining S1 and S2 from –θ to + θ Solution: Δ xP = Path difference at P = = S 1 A – S 2A 92 x2 x
As x → p, ΔXP → 0
e. in
No. of maxima = 5
je
No. of minima = 6
Coherent Sources
cr
ac
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Coherent sources are those which are derived from same one source (i.e., atomic oscillator) and the light emitted by them have same frequency. Amplitude of emitted wave may or may not be same. The phase relationship does not change with time. The initial phase difference at the time of emission between the waves from coherent sources is either zero or constant. Thus, coherent sources give out waves which have point to point phase correspondence (or relationship).
Production of Coherent Sources
By division of wave front
By division of amplitude
37
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By division of wave front: In this method the wave-front is divided into two or (a) more parts by the use of mirrors, lenses and prisms. The well known methods are Young’s double slit arrangement. Fresnel’s biprism and Lloyd’s single mirror etc.
e. in
(b) By division of amplitude: In this method the amplitude of the incoming beam is divided into two or more parts by partial reflection or refraction. These divided parts travel different paths and finally brought together to produce interference. This type of interference needs broad source of light. The common examples of such interference of light are the brilliant colours seen when a thin film of transparent material like soap bubble or thin film of kerosene oil spread on the surface of water is exposed to a broad source of light.
Ways of obtaining a pair of Coherent Sources
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(i) Double slit method: Light from a source ‘S’ is limited to a narrow beam (not a narrow beam but diverging source) with the help of a slit (Fig.). The emergent light is made to fall upon a screen containing two slits ‘S1’ and ‘S2’ placed symmetrically with respect to the slit. In that case both S1 and S2 are illuminated by the same wave-front. Therefore, the beams of light coming out from S1 and S2 have no phase difference.
Thus, ‘S1’ and ‘S2’ can be treated as the coherent sources. Young made use of them in his famous young’s double slit experiment.
38
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je
e. in
(ii) A source and its own virtual image: Light from a source ‘S’ is made to fall on a plane mirror M (Fig.). Point of observation P on a screen AB receives direct light as well as reflected light. For an observer reflected light appears to come from a source S (virtual image of S). So, interference at p takes place between waves coming from S and S’. Since S’ is not an independent source, being the virtual image of S, it will have the same phase as ‘S’.
ac
ki it
Hence, the two can be treated to be coherent source. Lloyd made use of this arrangement in Lloyd single mirror experiment.
cr
(iii) Bi-prism method : Light from a source S is made to fall on an assembly of two right angled prisms A and B joined base to base as shown in fig. S1 and S2 are the virtual images of S produced by refraction through prisms A and B respectively. Being virtual images of the same source, ‘S1’ and ‘S2’ have same phase and hence can be treated as the coherent sources. This type of arrangement is made use of in Frensel’s bi-prism experiment.
39
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e. in
Remarks
Two independent sources similar in every respect cannot act as coherent sources. The reason lies in the origin of light which is due to random emission of radiation from excited atom. When the atom is excited, say, by heating or electric discharge, etc, then the revolving electrons absorb energy and go to outer orbits which are unstable. Hence, soon the electrons spontaneously fall back to inner orbits thus emitting energy in the form of radiation. When the frequency of these radiations lies within the range of visible spectrum, we have the emission of light, which consists of a broken chain of wave-trains, accompanied with sudden and abrupt changes in phase occurring in very short intervals of time (of the order of 10–8 sec.). Consequently with two independent sources or with two separate portions of the same source, the phases of waves emanating from them will be changing independently of each other so that the phase difference between two such sources cannot be constant.
(ii)
Two identical laser sources can act as coherent sources.
cr
ac
ki it
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(i)
40
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cr
ac
ki it
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e. in
Young’s Double Slit Experiment
Δ x= Path difference at P = = S2P – S1A = d sin θ 41
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Positions of maxima Δ x= Path difference at = n d sin θ = n sin θ =
n d
je
Third Maxima
nD d
cr
Yn = nth maxima =
Second Maxima
ki it
y n D d
First Maxima
ac
Zero Order Maxima (Central Maxima)
e. in
n= Order of Maxima
42
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Positions of Minima Δ x= Path difference at (2n 1)
2
d sin θ = (2n 1)
n=0,1,2……… Second maxima
n=0,1,2……… First maxima
Second maxima
ac
First maxima
(2n 1) 2d
e. in
(2n 1) 2d
je
(2n 1) 2d
ki it
sin θ =
2
cr
Fringe width () This is defined as linear separation between two consecutive maxima or minima
yn yn1
D d
43
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Angular Fringe width () This is angular separation between two consecutive maxima or minima
D d
Geometry of Fringes For any fringe Δ x = constant
e. in
S2P – S1P = constant
ac
Intensity Distribution Is1 = Is2 = I (say)
ki it
je
The locus of any point which moves in such a way so that difference of its distance from two fixed point is constant, is known as hyperbola.
cr
Ip = 4 I cos2
Where = phase difference
λR > λV βR > βV
λ = wavelength of light in air λ1 = wavelength of light in medium =
44
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New fringe width (β’) =
New angular fringe width (ω) =
Displacement of fringes due to introduction of thin transparent medium in the path of one of the interfering beam.
e. in
Consider two coherent sources sending waves to reach P (Fig.) with a path difference x given by
je
d S 2P – S 1P = D
ki it
If ‘n’ is the order of bright fringe coinciding with vertical cross-wire of the field of view at P,
ac
y d =n× D
cr
Introduce a thin plate of transparent material (refractive index ) of thickness ‘t’ in one of the beams say S1P (fig.). Since velocity of light in the denser medium is lesser, light will take comparatively greater time to go from S1 to P in the presence of plate. Thus, net path difference between S2P and S1P will decrease. Net Path difference = S2P – [(S1P – t)air + tsheet] = S2P – [(S1P – t) + t] = (S2P – S1P) – ( -1)t
d = y D – ( -1)t x
yd ( 1)t n D 45
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yn = Position of nth maxima in presence of thin transparent sheet.
nD D ( 1)t n d d
For central maxima Δx = 0 yd ( 1)t 0 D
ac
ki it
je
e. in
D ( 1)t shift(s) d
cr
y
46
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Examples Example: A beam of light consisting of two wavelength 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment. (i)
Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å.
(ii)What is the least distance from the central maximum when the bright fringes due to both wavelengths coincide? The distance between the slit is 2 mm and the distance between the plane of the slits and the screen is 120 cm. 1 = 6500 Å = 6.5 × 10–7 m
e. in
Solution : Given,
2 = 5200 Å = 5.2 × 10–7 m
je
d = 0.2 cm = 2 × 10–3 m D = 120 cm = 1.2 m yn =
1
= 6.5 × 10–7 m
ac
Here,
n D d
ki it
(i) For nth bright spot
Ans.
cr
3 6.5 10 7 1.2 y3 = = 1.17 × 10–3 m 2 10 3
(ii)Since 2 < 1, fringe width for 2 is smaller. If two bright fringes due to 1 and 2 are to coincide, then minimum distance from the central spot will be where n th order bright spot due to 1 and (n + 1)th bright spot due to 2 coincide. n
(n 1)2 D 1 D = d d
n × 6.5 × 10–7 = (n + 1) × 5.2 × 10–7 y=
or
n=4
n D 4 6.5 10 7 1.2 = = 1.56 × 10–3 m Ans. d 2 10 3
47
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Example: In 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 is introduced in the path of one of interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift on the introduction of mica sheet. Calculate the wavelength of the monochromatic light used in the experiment. Solution : Shift ‘y’ in the fringe system is ( 1)t
e. in
y =
when distance between slits and screen is doubled,
Here,
ki it
( 1)t = 2
=
( 1)t 2
ac
= y
= 1.6,
cr
Given
je
= 2
=
t = 1.964 × 10–6 m
(1.6 1) 1.964 10 6 2
= 0.3 × 1.964 × 10–6 m = 5892 Å
Ans.
48
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Example: In a modified Young’s double slit experiment, monochromatic uniform and
10
–2 parallel beam of light of wavelength 6000 Å and intensity Wm incident normally on two circular operation A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength 6000 Å is placed in front of aperture A (fig.). Calculate the power (in watt) received all the focal spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume that 10% of the power received by each observer goes in the original direction and is brought to the focal spot.
Solution : Let I1 and I2 be the intensities at A and B 10
Wm–2
e. in
I1 = I2 =
r12
= × (0.001)2 = × 10–6 m2
Area of cross-section of aperture B, A2 =
r22
= × (0.001)2 = × 4 × 10–6 m2
je
Area of cross-section of aperture A, A1 =
ki it
Let P1 and P2 be the powers of incident radiations at A and B respectively. 10 6 P1 = 10 = 10–6 = 10–5 W
ac
10 6 P2 = 4 10 = 4 × 10–5 W
cr
Introduction of a transparent medium in one of the beams produces some path difference x. x = ( – 1) t
Here,
= 1.5
and t = 2000 Å x = (1.5 – 1) × 2000 Å = 0.5 × 2000 Å
or, Let
x = 10–7 m = phase difference between the two beams 2 = x
49
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or,
=
2 10 7 = 3 radian 6000 10 10
If a1 and a2 are the amplitudes of light from apertures A and B, net amplitude R at F is, R2 =
a12 a22 2 a1 a2 cos
Power = Intensity × Area of cross-section = I × A2
and
P = KR2 × A2 = KR2 P1 =
K a12
P2 =
K a22
Multiply equation (1) by K throughout
e. in
or,
... (1)
P = P1 P2 2 P1 P2 cos
ki it
or,
je
KR2 = K a12 K a22 2 K a1 K a2 cos
Substituting for P1, P2 and , we get
P = 7 × 10–5 W
cr
or,
ac
5 5 5 5 P = (10) 4 10 2 10 4 10 cos 3
Ans.
50
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Example: In fig. shown, S is a monochromatic point source emitting light of wavelength = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 nm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO. (i)
If the third intensity maximum occurs at the point P on the screen, find distance OP.
(ii)If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OP increase, decrease or remain the same?
cr
ac
ki it
je
e. in
Solution : (i) As shown in fig. each part of the lens will form image of S which will act as coherent sources.
From lens equation, we can write 1 1 v 15
or,
=
1 10
v = 30 cm m=
v u
=–2
d = 3 × 0.5 mm = 1.5 mm Also,
D = 1.30 – 0.30 = 1 m
Now, from the theory of interference the distance y of a point P on the screen is given by 51
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D y = d ( x ) and as point is third maximum x = 3 So,
D y = d (3 )
or,
7 y = 5 10 3 10 3 m = 1 mm Ans.
0.5 10
(iii) If gap between L1 and L2 is reduced then d will decrease. As
and OP = 3,
e. in
therefore OP will increase.
D d
ac
ki it
je
Example: A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8 Light of wavelength travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If = 648 nm, obtain the lease value of t for which the rays interfere constructively.
cr
Solution : The ray reflected from upper surface suffer a phase change of due to reflection, at denser media, so the condition of constructive interference for normal incidence is given by
2 t – 2 = n
or
2 t =
(2 n 1) 2 52
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For minimum value of t,
n=0
tmin = 4 = 90 nm
Ans.
ki it
je
e. in
Example: Fig. shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let BP0 AP0 = /3 and D >> . (a) Show that in this case d 2D /3 . (b) Show that the intensity at P0 is three times the intensity due to any of the three slits individually.
or, or,
BP0 – AP0 =
3
d sin =
3
d tan =
3
cr
or,
ac
Solution: (a)
d d /2 D 3
d=
(For small angle tan sin )
2 D 3
53
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(b)
xA/B = path difference between waves coming from A and B = 3
A/B = phase difference
2 2 = x A /B = 3
xB/C = d sin
Similarly,
2 3d / 2 = d D = 32dD =
B/C = 2
ki it
je
e. in
Now, phase diagram of the waves arriving at P0 is as shown below:
ac
Amplitude of resultant wave is given by
cr
A =
=
A 2 (2 A )2 2( A )(2 A )cos 120 3A
As intensity (I) A2
Intensity at P0 will be three times the intensity due to any of the three slits individually.
54
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Example: In a Young experiment the light source is at distance l1 = 20 m and l2 = 40 m from the slits. The light of wavelength = 500 nm is incident on slits separated at a distance 10 m. A screen is placed at a distance D = 2 m away from the slits as shown in fig. Find: The values of relative to the central line where maxima appear on the screen?
(b)
How many maxima will appear on the screen?
(c)
What should be minimum thickness of a slab of refractive index 1.5 be placed on the path of one of the ray so that minima occurs at C?
ki it
je
e. in
(a)
Solution: (a) The optical path difference between the beams arriving at P,
ac
x = (l1 – l2) + d sin
cr
The condition for maximum intensity is, x = n
Thus,
sin = d1 [x (l1 l2 )] = d1 [n (l1 l2 )] =
Hence,
=
1 [ n 500 10 9 20 10 6 ] = 2 n 1 40 10 10 6
s i n 1 2 n 1 40
Ans.
55
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|sin | 1
(b)
n –1 2 40 1 1
or,
– 20 (n – 4 20 n 60
or, Hence, (c)
Number of maxima = 60 – 20 = 40
At C, phase difference,
=
2 (l
2
Ans.
l1 ) =
500210 20 10 9
6
= 80
( – 1)t =
2
10 9 t = 2( 1) = 500 = 500 nm 2 0.5
Ans.
ki it
je
or,
e. in
Hence, maximum intensity will appear at C. For minimum intensity at C,
cr
ac
Example: Light of wavelength = 500 nm falls on two narrow slits placed a distance d = 50 × 104 cm apart, at an angle = 30 relative to the slits shown in fig. On the lower slit a transparent slab of thickness 0.1 mm and refractive index 3/2 is placed. The interference pattern is observed on a screen at a distance D = 2 m from the slits. Then calculate : (a) Position of the central maxima? (b) The order of minima closest to centre C of screen? (c) How many fringes will pass over C, if we remove the transparent slab from the lower slit?
56
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Solution:
x = 0
ac
For central maxima,
ki it
je
e. in
(a) Path difference, x = d sin + d sin – ( – 1) t
sin =
( 1)t s in d
cr
=
(3/2 1)(0.1) s in 30 50 10 3
1 = 2 = 30°
(b)
At C,
Therefore,
Ans.
= 0°, x = d sin – ( – 1) t = (50 × 10–3)
12 – (3/2 – 1) (0.1) 57
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= 0.025 – 0.05 = – 0.025 mm x = n,
Substituting
n = x =
We get
0.025 = – 50 500 10 6
Hence, at C, there will be maxima. Therefore, closed to C order of minima is 49. Ans. (c) Number of fringes shifted upwards =
( 1)t (3/2 1)(0.1) = = 100 500 10 6
Ans.
cr
ac
ki it
je
e. in
Example: Consider the situation shown in fig. The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength . The separation between the slits is d. The light transmitted by the slits falls on a screen 1 placed at a distance D from the slits. The slit S3 is at the central line and the slits S4 is at a distance z from S3. Another screen 2 is placed a further distance D away from 1. Find the ratio of the maximum to minimum intensity observed on 2 if z is equal to
D
(a) 2 d
D
(b) 4 d
Solution: (a)
z= D 2d
58
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Let I is intensity due to slits S1 and S2 1 is given by IP =
4 I cos 2
At slit S3,
=0
I S3
2
= 4I
dz
x = D = 2 =
I S4 = 0
or
e. in
At slit S4,
Now on screen 2 I S3
I S4
Imin =
I S3
I S4
Im ax I m in
D 4d
Ans.
=1
= 4I
cr
I S3
At slit S4,
2 = 4I
ac
z=
2 = 4I
je
Imax =
ki it
(b)
Further, intensity at any point on
1.
x =
dz D
=
4
=
=
2 4
I S4
=
Imax =
I S3
I S4
2 =
4 I 2I
=
Similarly,
Imin =
I S3
I S4
2 =
4 I 2I
= I 2
Im ax I m in
2
4 I cos 2 2 I 4
=
22 22
2
2
2
I 2 2
2
2
2
Ans. 59
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Example: A convex lens of focal length ƒ is used as a simple microscope. Find its magnifying power if the final image is at the distance of distinct vision. Find also the magnification and the distance of the object ƒ = 5 cm, D = 25 cm. Solution : The object is within the focal distance
M=
i O
=
1 25 5
Magnification =
=6
A B AB
25 6 25
25 u1
25 5 25 5
=
125 30
=
25 6
cm
= 6 Ans.
ki it
Magnification =
=
v1 u1
=
=
ac
1 D ƒ
v1 ƒ v1 ƒ
u1 =
AB D
cr
and O =
AB u1
=
e. in
A B v1
je
i =
60
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Example: The focal lengths of the objective and the eyepiece of a compound microscope for 4 cm and 6 cm respectively. If an object is placed at a distance of 6 cm in front of the objective, what is the magnification produced by the microscope? Distance of distinct vision is 25 cm.
Ma gn ifica t ion of m icr os cop e
Ma gn ifica t ion pr odu ced by t h e object ive
m=
v 1 D u ƒe
×
An gu la r m a gn ifica t ion pr odu ced by t h e eyepiece
ac
=
ki it
je
e. in
Solution :
cr
u = 6 cm, ƒO = 4 cm, ƒe = 6 cm, D = 25 cm v=
m=
u ƒO u ƒO
=
64 2
12 1 25 6 6
= 12 cm
= 2 316 = 10.33
Ans.
61
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Example: A telescope has an objective of focal length 50 cm and an eye piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate: (i) The separation between the objective and the eyepiece. (ii) The magnification produced. Solution : Let AB be the position of the object u = 200 cm,
ƒ = 50 cm
e. in
The image AB formed by the objective is a distance v from it uƒ
200 50 v = u ƒ = 200 = 50
1000 200 15 = 3 cm
l 200 3
, where l is the separation between the lenses. The images (final) distance is
ki it
u =
je
This serves as an object for the eyepiece. The distance between AB and the eyepiece is v = 25 cm from the eyepiece and ƒ = 5 cm. v ƒ v ƒ
=
25 5 25 5
ac
u =
l=
25 200 6 3
=
425 6
=
25 6
cm
= 70.83 cm
cr
v v 200 1 6 Total magnification mO × me = u u = 3 200 25 25 = 2 Ans.
62
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