TEACHING NOTES WAVE
OPTICS
Coherent and Incoherent Sources Why do we not commonly see interference effects with visible light? With light from a source such as the Sun, an incandescent bulb, or a fluorescent bulb, we do not see regions of constructive and destructive interference; rather, the intensity at any point is the sum of the intensities due to the individual waves. Light from anyone of these sources is, at the atomic level, emitted by a vast number of independent sources. Waves from independent sources are incoherent; they do not maintain a fixed phase relationship with each other. We cannot accurately predict the phase (for instance, whether the wave is at a maximum or at a zero) at one point given the phase at another point. Incoherent waves have rapidly fluctuating phase relationships. The result is an averaging out of interference effects, so that the total intensity (or power per unit area) is just the sum of the intensities of the individual waves. Only the superposition of coherent waves produces interference. Coherent waves must be locked in with a fixed phase relationship. Coherent and incoherent waves are idealized extremes; all real waves fall somewhere between the extremes. The light emitted by a laser can be highly coherent-two points in the beam can be coherent even if sep arated by as much as several kilometers. Light from a distant point source (such as a star other than the Sun) has some degree of coherence. Thomas Young (1773-1829) performed the first visible-light interference experiments using a clever technique to obtain two coherent light sources from a single source. When a single narrow slit is illuminated, the light wave that passes through the slit diffracts or spreads out. The single slit acts as a single coherent source to illuminate two other slits. These two other slits then act as sources of coherent light for interference.
INTENSITY OF TWO SOURCE INTERFERENCE We now obtain an expression for the distribution of intensity of two coherent sources that are in phase. The wave function in this case is the electric field. We assume that the slits are narrow enough for diffraction to spread light from each slit uniformly over the screen. Thus, the amplitude of the fields any point on the screen will be equal. At a given point of the screen the fields due to S1 and S2 are E1 = E0 win (t) ; E2 = E0 sin (t + ) where the phase difference depends on the path difference = r2 – r1. Since one wavelength corresponding to a phase change of 2, a distance corresponds to a phase change given by / 2 = / . If the screen is far from the slits, d sin (see the ref. figure for YDSE), therefore 2 2d sin = The resultant field is found from the principle of superposition : E = E1 + E2 = E0 sin (t) + E0 sin (t + )
=
Page # 1
By using the trigonometric identity sin A + sin B = 2 sin [(A + B)/2] cos [(A – B)/2], we obtain E = 2E0 cos sin t 2 2 The amplitude of the resultant wave is 2E0 cos (/ 2). The intensity of a wave is proportional to the square of the amplitude, so from equation wave have I = 4I0 cos2 cos2 2 where I0 E02 is the intensity due to a single source. This function is plotted in fig. of pt (iv). The maxima occur when = 0, 2, 4, ... = 2m. At these points I = 4I0 ; that is the intensity is four times that of a single source. The minima (I = 0) occur when = , 3, 5, ....... = (2m + 1).
(i)
If amplitudes of waves arriving at point P on the screen are different then resultant intensity is given by I = I1 + I2 + 2 I1I 2 cos Also,
=
, 2
Imax =
I1 I 2 ,
Imin
I1 I 2
when cos = 1
2
when cos = –1 2
2 2 I1 I 2 I max A1 A 2 r 1 I min = I1 I 2 = A1 A 2 = r 1
(ii)
A1 I1 r= A I 2 2
where (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
Id Iav =
0 2
d
2
1 (I1 I 2 2 I1I 2 cos I1 I 2 2 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.
Page # 2
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 figure.
Intensity (I)
(iv)
I max
I1 I 2
2
I av I1 I 2 O /2
I min
I1 I 2
2
Path difference (x)
(iv)
Path difference (x) and phase difference () are related as given below : path difference = 2 phase difference or
x =
2
YOUNG’S DOUBLE SLIT EXPERIMENT (YDSE) The experiment set up for Young’s double slit experiment is shown in figure. Light after passing through a pin hole ‘S’ is allowed to fall on thin slits ‘S1’ and ‘S2’ placed symmetrically w.r.t. ‘S’. A screen is placed at a distance ‘D’ from S1 and S2. P S1 y
S1
S2
S
dQ
O
S2
x
Monochromatic Source
D
Screen
(D >> d) Geometric construction for describing Young’s double-slit experiment
Let ‘P’ be the point, at which we want to investigate the intensity. Two rays S1P and S2P starting from S1 and S2 reach P and interfere with each other. If x is the path difference between two rays, y dy For small angle sin tan D D This equation assumes that S1P and S2P are parallel, which is approximately true because D is much greater than d. Maxima : Point ‘P’ will be a bright spot if the path difference x is integral multiple of .
x = S2P – S1P d sin
(a)
yn =
nD d
where, n = 0, 1, 2, 3, .......
Thus, bright spots are obtained at distances, 0, (b)
Minima :
D 2D 3D , , ....... from O. d d d
Point ‘P’ will be a dark spot if the path difference ‘x’ is an odd multiple of
. 2
Page # 3
yd ( 2n 1) = D 2
i.e., if
yn =
( 2n 1)D where, n = 0, 1, 2, 3, ........ 2d
Thus, dark spots are obtained at distances, (c)
D 3 D 5 D , , ....... from O. 2d 2d 2d
Fringe width () It is the distance between two consecutive bright or dark fringes.
Let yn and yn – 1 respectively, be the distances of nth and (n – 1)th bright fringe from O, = [yn – yn–1] = n
D D D (n – 1) = (n – n + 1) d d d
D d Similarly, it can be proved that distance between two consecutive dark fringes, is given by
or
=
D d = = D / d Hence, the bright and dark fringes are equally spaced.
=
[Home Work : HCV : 1 to 11 Sheet Ex.1 : 1 Q. Bank : 1, 3
Ex. 3 : 1, 6 AR : 2
]
Remarks : (i) If whole apparatus is immersed in liquid of refractive index µ then,
D i.e., fringes width decreases d Some times in numerical problems, angular fringe width () is given which is defined as angular separation between two consecutive maxima or minima =
(ii)
D d In medium, other than air or vacuum,
=
=
d Page # 4
yd is valid when angular position of maxima or minima is less than . However x = d sin is D 6 valid for larger values of provided d << D.
(iii)
x =
(iv)
Central bright fringe (CBF) is a point on screen where path difference is zero. In above case CBF is formed at O. But in many situation it may not be located symmetrically w.r.t. slits.
(v)
If white light is used instead of monochromatic light then, interference pattern consists of white central bright fringe surrounded by few coloured fringes and then uniform illumination due to overlapping of interference pattern on each wavelength.
(vi)
If the interference experiment is performed with bichromatic light, the bright fringes of two wavelength will be coincident for the first time under following condition. Y = n ()Longer = (n + 1) Shorter or nLonger = (n + 1) Shorter
(vii)
In many numerical problems we have to calculate number of maxima or minima. We know that for maximum. sin =
n
n d
or
d 1
n=
d sin
( sin 1)
d nlighest = Lighest order of maxima on one side. d works out to be 2.3 so, permissible values of n are 0, ± 1, ± 2. Hence, total 5 maxima will be obtained on screen.
Suppose in some question
Ex.
In YDSE experiment the distance between slits is d = 0.25 cm and the distance of screen D = 120 cm from slits. If the wavelength of light used is = 6000Å and I0 is the intensity of central maximum, at what distance from the centre of the intensity will be (A) 4.8 × 10–5 m
Ex.
[Sol.
(B*) 7.2 × 10–5 m
I0 ? 2 (C) 3.6 × 10–5 m
(D) 5.4 × 10–5m
In the figure, if a parallel beam of white light is incident on the plane of the slits then the distance of the white spot on the screen from O is [assume d << D, << d] (A) 0 (B) d/2 (C*) d/3 (D) d/6 White spot will be central max., where path diff. is zero
OP = 5d/6 – d/2 = d/3 Ans.] Page # 5
SHAPE OF INTERFERENCE FRINGES IN YDSE We discuss the shape of fringes when two pinholes are used instead of the two slits in YDSE. Fringes are locus of points which move in such a way that its path difference from the two slits remains constant. y
P x
S1 • x
O S2•
S2P – S1P = D = constant
S1
If = ± , the fringe represents 1st minima. 2
(A) (B) (C)
Y
........(1)
= 3 = 2 = X
= 3 = If = ± it represents 2nd minima 2 = S2 If = 0 it represents central maxima, If = ± , it represents 1st maxima etc. Equation (1) represents a hyperbola with its two foci at S1 and S2 The interference pattern which we get on screen is the section of hyperboloid of revolution when we revolve the hyperbola about the axis S1S2. If the screen is to the X-axis, i.e. in the YZ plane, as is generally the case, fringes are hyperbolic with a straight central section. If the screen is in the XY plane, again fringes are hyperbolic. If screen is to Y-axis (along S1S2) i.e. in the XZ plane, fringes are concentric circles with center on the
axis S1S2 ; the central fringe is bright if S1S2 = n and dark if S1S2 = (2n – 1)
. 2
Y
Z
Y
Y
A
X
X
B
[Home Work :HCV : 29 to 32 Sheet Ex.1 : 2 to 6 Ex.2 : 1 Q Bank MCQ : 6, 7, 8, 29
C
Ex.3 : 7, 12, 13 ]
{tell students to attempt from the book or can be given as HW} HCV-Ex. 33 Page # 6
Ex.
Consider the situation shown in figure. 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
S1
S4 z
d
S3 S2 1 2 D
D
D D (b) 2d 4d [Sol.(a) Let I is intensity due to slits S1 and S2 on screen S1. Further, intensity at any point on screen 1 is given by
(a)
2
I P = 4I cos2
At slit S3,
=0
IS3 = 4I
At slit S4,
x =
=
Now on screen 2
Imax =
S3
IS4
Imin
S3
I S4
dz = D 2
IS 4 = 0
I = I
2
2
= 4I = 4I
I max I min = 1 Ans
(b)
z=
D 4d
IS3 = 4I At slit S4,
x =
dz = D 4
2 × = 4 2
=
IS4 = 4I cos2 = 2I 4
Page # 7
IS3
= I (2 +
2 )2
IS3
Imax =
Similarly,
Imin =
I S4
=
4I
2I
I S4
=
4I
2I
2 2 I max = 2 2 I min
2
2
2
2
=I 2
2
2
2
Ans.
HCV-Ex. 28 Ex. Figure 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. C d B d A D
P0
C d
B
Sol.
d A
D
BP0 – AP0 = or or
P0
3
3 d tan = 3
d sin =
(For small angle tan sin )
d/2 = d D 3
or
d=
(b)
xA/B = path difference between waves coming from A and B =
A/B = phase difference
2F 3 3
2 2 xA/B = 3 Similarly, xB/C = d sin
=
Page # 8
3d / 2 3d 2 = =d =1 D 2D B/C = 2 Now, phase diagram of the waves arriving at P0 is as shown below : A
2A
A
120°
120° A
A
Amplitude of resultant wave is given by A’ =
A 2 (2A ) 2 2( A )( 2A ) cos 120 =
3A
As intensity (I) A2 Intensity at P0 will be three times the intensity due to any of the three slits individually. [Home Work : HCV : 20, 25, 26, 27, 28, 33, 34 Sheet Ex-1 : 11, 12, 13 Ex.3 : 4, 14, 18 Q. Bank : Single correct 4, 5, 9, 10, 11, 13, 14, 15, 16, 24 MCQ 2, 3, 4, 10, 11, 12, 13 AR 3 ] GEOMETRICAL PATH AND OPTICAL PATH Actual distance travelled by light in a medium is called geometrical path (x). Consider a light wave given by the equation. E = E0 sin (t – kx + ) If the light travels by x, its phase changes by kx =
x, where , the frequency of light does not v
depend on the medium, but v, the speed of light depends on the medium as v =
c .
Consequently, change in phase, (µx) C It is clear that a wave travelling a distance x in a medium of refractive index µ suffers the same phase change as when it travels a distance µx in vacuum. i.e. a path length of x in medium of refractive index µ is equivalent to a path length of µx in vacuum.
= kx =
The quantity µx is called the optical path length of light, xopt. And in terms of optical path length, phase difference would be given by, 2 xopt = xopt C 0 where 0 = wavelength of light in vacuum. However in terms of the geometrical path length x,
=
=
2 (µx) = x C
0 . where = wavelength of light in the medium
Page # 9
1.
Displacement of fringe on introduction of a glass slab in the path of the light coming out of the slits. P
t S1
O’
d O
S2 D
On intrudction of the thin glass-slab of thickness t and refractive index µ, the optical path of the ray S1P increases by t(µ – 1). Now the path difference between waves coming from S1 and S2 at any point P is p = S2P – (S1P + t(µ – 1)) = (S2P – S1P) – t(µ – 1) p = d sin – t (µ – 1) if d << D yd – t (µ – 1) if y << D as well. D for central bright fringe, p = 0 yd = t(µ – 1). D
and
p
=
D = (µ – 1) t d The whole fringe pattern gets shifted by the same distance
y = OO’ = (µ – 1)t
D B = (µ – 1)t d Notice that this shift is in the direction of the slit before which the glass slab is placed. If the glass slab is placed before the upper slit, the fringe pattern gets shifted upwards and if the glass slab is placed before the lower slit the fringe pattern gets shifted downwards.
= (µ – 1)t
*
Ex.
A young's double slit experiment is conducted in water (1) as shown in the figure, and a glass plate of thickness t and refractive index 2 is placed in the path of S2. Wavelength of light in water is . Find the magnitude of the phase difference between waves coming from S1 and S2 at 'O'. 2 2 (A*) 1 t 1
1 2 (B) 1 t 2
(C) ( 2 1 ) t
[Home Work :HCV : 12 to 19 Sheet Ex- : 7, 8 Ex.2 : 2, 3, 4 Q.Bank Single correct - 17 to 23 MCQ 5, 6, 8, 9, 14 AR 1
2
(D) ( 2 1) t
2
Ex.3 : 2, 8, 19
]
Page # 10
The phase change on reflection A ray of light is incident on air-water interface ; let the amplitude reflection and transmission coefficients be r1 and t1 respectively. The amplitudes of reflected and transmitted waves are ar1 and at1 respectively. From the principle of reversibility of light, the system retraces its whole previous motion. The wave of amplitude ar1 gives a reflected wave of amplitude ar1t1. The wave of amplitude at1
r1,t2
at 1t2
r2,t1
gives a reflected wave of amplitude ar2t1 and transmitted wave amplitude at1t2. So ar12 + at1t2 = a t1t2 = 1 – t12 ... (1) Further, the waves of amplitudes at1r2 and ar1t1 must cancel each other. at1r2 + ar1t1 = 0 r 2 = – r1 ... (2) equation shows a difference of phase of between the two cases ; a reversal of sign means a displacement in the opposite sense. If there is no change of phase on reflection from above, there must be a phase change of from below and vice-versa. When light gets reflected from a denser medium there is an abrupt phase change of ; no phase change occurs when reflection takes place from rarer medium. Thin film interference When light passes the boundary between two transparent media, some light is reflected at the boundary. As shown in the figure some light is reflected from first interface and some from second interface. If we consider a monochromatic incident light the two reflected waves are also monochromatic and coherent because they arise from the same monochromatic incident light wave via amplitude division. These waves interfere, since they are superposed along the same normal line. The phase difference between two interfering waves is due to : (1) Optical path difference (due to distances travelled), (2) Reflection from a denser medium.
The second factor is irrelevant for reflection at rarer medium. Three situations may arise : (1) Neither wave experiences a phase change upon reflection. (2) Both the waves suffer a phase change upon reflection. In either of these two cases the phase change due to reflection is irrelevant ; no difference in phase results due to reflection. In either of these cases phase change is determined solely from optical path difference. Condition for constructive interference :
Page # 11
2t = m Condition for destructive interference : 1 2t = m 2 where m = 0, 1, 2, ....... (3) One of the reflected waves experiences a phase change of radian upon reflection and the other waves does not. It is material which wave suffers a phase change ; the conclusions in the previous case are first reversed. Condition for destructive interference : 2t = m Condition for constructive interference : 1 2t = m 2 where m = 0, 1, 2, ....... Ex.
A light ray is incident normal to a thin layer of glass. Given the figure, what is the minimum thickness of the glass that gives the reflected light an orangish color ( air = 600 nm)? (A) 50 nm (B*) 100 nm (D) 200 nm (E) 500 nm
[Sol.
(C) 150 nm
For reflected light to have orangish color, rays from A, C, E must be out of phase for l = 600 nm or = (2n + 1) Reflected light or i.e.
or
2gt (2n + 1)
2
air (rare) glass A
C
E t
t = (2n 1) 4 g
(dense) (rare) water B
D
tmin = 4 = 100 nm ] g
Transmitted light
(OPTIONAL) INTERFERENCE DUE TO REFLECTED LIGHT : Consider a transparent film of thickness t and refractive index µ. A ray SA incident on the upper surface of the film is partly reflected along AR1 and partly refracted along AB. At B part of it is reflected along BC and finally emerges out along CR2. The difference in path between the two rays. AR1 and CR2 is calculated as given below : Let CN and BM be perpendicular to AR1 and AC. As the paths of the rays AR1 and CR2 beyond CN are equal. The path difference between them is x = Path ABC in film – Path AN in air = µ(AB + BC) – AN = 2µ AB – AN S
N R1
R2
i i
C
Air
M
A r
t r B
Air T1
Page # 12
Now,
AB = BC =
and
AN = =2
AB × BM = BM sec r = t sec r BM
AN · AC = AC sin i = 2 AM sin i AC
sin i AM sin r BM sin i = 2 (tan r) t BM sin r
sin 2 r = 2µt sec r sin2 r cos r Then, = 2µAB – AN = 2µt sec r – 2 µt sec r sin2 r = 2 µt sec r (1 – sin2 r) = 2 µt cos r The ray AR1 having suffered a reflection at the surface of denser medium undergoes a phase change or
= 2µt
. 2 At B the reflection takes place when the ray is going from a denser to rarer medium and there is no phase change. Hence, the effective path difference between AR1 and CR2 is given by
path diff. of
2 If the path difference x = n where n = 0, 1, 2, 3, 4 etc., constructive interference takes place and the film appears bright.
Path Diff. (x) = 2 µt cos r – (i)
= n 2
2µt cos r –
or
2µt cos r = (2n + 1)
(ii)
If the path difference x = (2n + 1)
2
where n = 0, 1, 2, ........... etc., destructive interference 2 takes place and the film appears dark.
2µt cos r –
= (2n + 1) or 2µt cos r = n 2 2
Remarks : (i) If the thickness of the film is very small as compared to the wavelength of light used, so that 2µt cos r can be neglected, then the total path difference between AR1 and CR2 will reduce to
. Thus two rays will 2
interfere destructively and darkness will result. (ii)
It should be remembered that the interference pattern will not be perfect because the intensities of the ray AR1 and CR2 will not be the same and their amplitudes are
Page # 13
Fringes of equal thickness Soap bubbles and oil films on a road do not have uniform thickness of the film at any given point determines whether the reflected light has a maximum or minimum intensity. When white light is used, each wavelength has its own fringe pattern. At a given point of the film, one wavelength may be enhanced and / or another wavelength suppressed. This is the source of the colors in soap bubbles an oil films on the road. A wedge-shaped film of air may be produced by placing a sheet of paper or a hair between the ends of two glass plates, as in fig. With flat plates, one sees a series of bright and dark bands, each characteristic of a particular thickness. If the plates are not flat, the fringes are not straight ; each is locus of points with the same thickness. If one plate is known to be flat, the fringes display the irregularities of the other, as shown in figure. The pattern shows where the plate needs to be polished for it to be made “optically flat.”
Ex.
Sol.
A wedge-shaped film of air is produced by placing a fine wire of diameter D between the ends of two flat glass plates of length L = 20 cm, as in fig. When the air film is illuminated with light of wavelength = 550 nm, there are 12 dark fringes per centimeter. Find D. A indicated in fig. only one of the reflected ray suffers a phase inversion. At the thin end of the wedge, where the thickness is less than /4, the two rays interfere destructively. This region is dark in the reflected light. The condition for destructive interference in the reflected light is 2t = m m = 0, 1, 2, ..... the change in thickness between adjacent dark fringes is t = / 2. The horizontal spacing between fringes d = 1/12 cm = 8.3 × 10–4 m. From figure we see that D / L = t / , so
(5.5 10 7 m) (0.2m) L = = 16.6 10 4 m 2d Thus
D = 6.6 × 10–5 m
Newton’s Rings When a lens with a large radius of curvature is placed on a flat plate, as to fig. a thin film of air is formed. When the film is illuminated with monochromatic light, circular fringes, called newton’s rings, can be with the unaided eye or with a low power microscope (figure). An important feature of Newton’s rings is the dark central spot. Newton tried polishing the surfaces to get rid of it, The dark spot was also initially puzzling to Young. It implied that the light wave suffers a phase inversion on reflection at a medium with a higher refractive index. Young tested this idea by placing oil of sassafras between a lens of crown glass and a plate of flint glass. The refractive index of the oil is between the values for these two glasses. Since both reflections occur at a medium with a higher refractive index, they should both suffer a phase inversion and therefore be in phase be in phase. This is precisely what happened : The central spot became bright and undoubtedly gave Young much satisfaction.
Page # 14
Ex. Sol.
In an experiment on Newton’s rings the light has a wavelength of 600 nm. The lens has a refractive index of 1.5 and a radius of curvature of 2.5 m. Find the radius of the 5th bright fringe. If R is the radius of curvature of the lens, then from fig. we see that r2 = R2 – (R – t)2, where r is the radius of a fringe and t is the thickness of the film. Since t is very small, we may drop terms in t2 to obtain. r2 2Rt ... (i) In order to find r, we must first find t. The condition for a bright fringe is 1 2t = m F ... (ii) 2 We note that n = 1 for the air film (the index for the glass is irrelevant) and that m = 4 for the fifth bright fringe. Thus, from (ii)
t=
(4.5) (6 10 7 ) = 1.35 × 10–6 m 2
Substituting this into (i), we find r = 2Rt = 2.6 × 10–3 m. [ Home Work : Sheet Ex.1 : Ex.2 : Q.Bank :
HCV : 35, 36, 37, 38 14 to 17 8, 9 Single correct - 25, 26, 27 MCQ - 7
]
HUYGEN’S CONSTRUCTION (This matter is from NCERT) Huygens, the Dutch physicist and astronomer of the seventeenth century, gave a beautiful geometrical description of wave propagation. We can guess that he must have seen water waves many times in the canals of his native place Holland. A stick placed in water and oscillated up and down becomes a source of waves. Since the surface of water is two dimensional, the resulting wavefronts would be circles instead of spheres. At each point on such a circle, the water level moves up and down. Huygen’s idea is that we can think of every such oscillating point on a wavefront as a new source of waves. According to Huygen’s principle, what we observe is the result of adding up the waves from all these different sources. These are called secondary wave or wavelets. Huygen’s principle is illustrated in (figure 10.1) in the simple case of a plane wave. (i) Consider a plane wave passing through a thin prism. Clearly, the portion of the incoming wavefront which travels through the greatest thickness of glass has been delayed the most. Since light travels more slowly in glass. This explains the tilt in the emerging wavefront. (ii) Similarly, the central part of an incident plane wave traverses the thickest portion of a convex lens and is delayed the most. The emerging wavefront has a depression at the centre. It is spherical and converges to a focus. Page # 15
(iii)
(iv)
A concave mirror produces a similar effect. The centre of the wavefront has to travel a greater distance before and after getting reflected, when compared to the edge. This again produces a converging spherical wavefront. Cocave lenses and convex mirror can be understood from time delay arguments in a similar manner. One interesting property which is obvious from the pictures of wavefronts is that the total time taken from a point on the object to the corresponding point on the image is the same measured along any rays going through the centre are shorter. But because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens.
[Home Work :Sheet : Q. bank :
Ex.3 : 3, 15, 16, 17 single correct - 2, 12, 30
]
Page # 16
