Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
9 | 1239-1242 | (b)
No Energy carried by a wave depends on the amplitude of the
wave, not on the speed of wave propagation (c)
For a given frequency, intensity of light in the photon picture is
determined by the number of photons crossing an unit area per
unit time (a)
(b)
FIGURE 10 |
9 | 1240-1243 | Energy carried by a wave depends on the amplitude of the
wave, not on the speed of wave propagation (c)
For a given frequency, intensity of light in the photon picture is
determined by the number of photons crossing an unit area per
unit time (a)
(b)
FIGURE 10 8 (a) Two needles oscillating in
phase in water represent two coherent sources |
9 | 1241-1244 | (c)
For a given frequency, intensity of light in the photon picture is
determined by the number of photons crossing an unit area per
unit time (a)
(b)
FIGURE 10 8 (a) Two needles oscillating in
phase in water represent two coherent sources (b) The pattern of displacement of water
molecules at an instant on the surface of water
showing nodal N (no displacement) and
antinodal A (maximum displacement) lines |
9 | 1242-1245 | (a)
(b)
FIGURE 10 8 (a) Two needles oscillating in
phase in water represent two coherent sources (b) The pattern of displacement of water
molecules at an instant on the surface of water
showing nodal N (no displacement) and
antinodal A (maximum displacement) lines Rationalised 2023-24
263
Wave Optics
Since the distances S1 P and S2 P are equal, waves from S1 and S2 will
take the same time to travel to the point P and waves that emanate from
S1 and S2 in phase will also arrive, at the point P, in phase |
9 | 1243-1246 | 8 (a) Two needles oscillating in
phase in water represent two coherent sources (b) The pattern of displacement of water
molecules at an instant on the surface of water
showing nodal N (no displacement) and
antinodal A (maximum displacement) lines Rationalised 2023-24
263
Wave Optics
Since the distances S1 P and S2 P are equal, waves from S1 and S2 will
take the same time to travel to the point P and waves that emanate from
S1 and S2 in phase will also arrive, at the point P, in phase Thus, if the displacement produced by the source S1 at the point P is
given by
y1 = a cos wt
then, the displacement produced by the source S2 (at the point P) will
also be given by
y2 = a cos wt
Thus, the resultant of displacement at P would be given by
y = y1 + y2 = 2 a cos wt
Since the intensity is proportional to the square of the amplitude, the
resultant intensity will be given by
I = 4 I0
where I0 represents the intensity produced by each one of the individual
sources; I0 is proportional to a2 |
9 | 1244-1247 | (b) The pattern of displacement of water
molecules at an instant on the surface of water
showing nodal N (no displacement) and
antinodal A (maximum displacement) lines Rationalised 2023-24
263
Wave Optics
Since the distances S1 P and S2 P are equal, waves from S1 and S2 will
take the same time to travel to the point P and waves that emanate from
S1 and S2 in phase will also arrive, at the point P, in phase Thus, if the displacement produced by the source S1 at the point P is
given by
y1 = a cos wt
then, the displacement produced by the source S2 (at the point P) will
also be given by
y2 = a cos wt
Thus, the resultant of displacement at P would be given by
y = y1 + y2 = 2 a cos wt
Since the intensity is proportional to the square of the amplitude, the
resultant intensity will be given by
I = 4 I0
where I0 represents the intensity produced by each one of the individual
sources; I0 is proportional to a2 In fact at any point on the perpendicular
bisector of S1S2, the intensity will be 4I0 |
9 | 1245-1248 | Rationalised 2023-24
263
Wave Optics
Since the distances S1 P and S2 P are equal, waves from S1 and S2 will
take the same time to travel to the point P and waves that emanate from
S1 and S2 in phase will also arrive, at the point P, in phase Thus, if the displacement produced by the source S1 at the point P is
given by
y1 = a cos wt
then, the displacement produced by the source S2 (at the point P) will
also be given by
y2 = a cos wt
Thus, the resultant of displacement at P would be given by
y = y1 + y2 = 2 a cos wt
Since the intensity is proportional to the square of the amplitude, the
resultant intensity will be given by
I = 4 I0
where I0 represents the intensity produced by each one of the individual
sources; I0 is proportional to a2 In fact at any point on the perpendicular
bisector of S1S2, the intensity will be 4I0 The two sources are said to
interfere constructively and we have what is referred to as constructive
interference |
9 | 1246-1249 | Thus, if the displacement produced by the source S1 at the point P is
given by
y1 = a cos wt
then, the displacement produced by the source S2 (at the point P) will
also be given by
y2 = a cos wt
Thus, the resultant of displacement at P would be given by
y = y1 + y2 = 2 a cos wt
Since the intensity is proportional to the square of the amplitude, the
resultant intensity will be given by
I = 4 I0
where I0 represents the intensity produced by each one of the individual
sources; I0 is proportional to a2 In fact at any point on the perpendicular
bisector of S1S2, the intensity will be 4I0 The two sources are said to
interfere constructively and we have what is referred to as constructive
interference We next consider a point Q [Fig |
9 | 1247-1250 | In fact at any point on the perpendicular
bisector of S1S2, the intensity will be 4I0 The two sources are said to
interfere constructively and we have what is referred to as constructive
interference We next consider a point Q [Fig 10 |
9 | 1248-1251 | The two sources are said to
interfere constructively and we have what is referred to as constructive
interference We next consider a point Q [Fig 10 9(a)]
for which
S2Q –S1Q = 2l
The waves emanating from S1 will arrive exactly two cycles earlier
than the waves from S2 and will again be in phase [Fig |
9 | 1249-1252 | We next consider a point Q [Fig 10 9(a)]
for which
S2Q –S1Q = 2l
The waves emanating from S1 will arrive exactly two cycles earlier
than the waves from S2 and will again be in phase [Fig 10 |
9 | 1250-1253 | 10 9(a)]
for which
S2Q –S1Q = 2l
The waves emanating from S1 will arrive exactly two cycles earlier
than the waves from S2 and will again be in phase [Fig 10 9(a)] |
9 | 1251-1254 | 9(a)]
for which
S2Q –S1Q = 2l
The waves emanating from S1 will arrive exactly two cycles earlier
than the waves from S2 and will again be in phase [Fig 10 9(a)] Thus, if
the displacement produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt – 4p) = a cos wt
where we have used the fact that a path difference of 2l corresponds to a
phase difference of 4p |
9 | 1252-1255 | 10 9(a)] Thus, if
the displacement produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt – 4p) = a cos wt
where we have used the fact that a path difference of 2l corresponds to a
phase difference of 4p The two displacements are once again in phase
and the intensity will again be 4 I0 giving rise to constructive interference |
9 | 1253-1256 | 9(a)] Thus, if
the displacement produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt – 4p) = a cos wt
where we have used the fact that a path difference of 2l corresponds to a
phase difference of 4p The two displacements are once again in phase
and the intensity will again be 4 I0 giving rise to constructive interference In the above analysis we have assumed that the distances S1Q and S2Q
are much greater than d (which represents the distance between S1 and
S2) so that although S1Q and S2Q are not equal, the amplitudes of the
displacement produced by each wave are very nearly the same |
9 | 1254-1257 | Thus, if
the displacement produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt – 4p) = a cos wt
where we have used the fact that a path difference of 2l corresponds to a
phase difference of 4p The two displacements are once again in phase
and the intensity will again be 4 I0 giving rise to constructive interference In the above analysis we have assumed that the distances S1Q and S2Q
are much greater than d (which represents the distance between S1 and
S2) so that although S1Q and S2Q are not equal, the amplitudes of the
displacement produced by each wave are very nearly the same We next consider a point R [Fig |
9 | 1255-1258 | The two displacements are once again in phase
and the intensity will again be 4 I0 giving rise to constructive interference In the above analysis we have assumed that the distances S1Q and S2Q
are much greater than d (which represents the distance between S1 and
S2) so that although S1Q and S2Q are not equal, the amplitudes of the
displacement produced by each wave are very nearly the same We next consider a point R [Fig 10 |
9 | 1256-1259 | In the above analysis we have assumed that the distances S1Q and S2Q
are much greater than d (which represents the distance between S1 and
S2) so that although S1Q and S2Q are not equal, the amplitudes of the
displacement produced by each wave are very nearly the same We next consider a point R [Fig 10 9(b)] for which
S2R – S1R = –2 |
9 | 1257-1260 | We next consider a point R [Fig 10 9(b)] for which
S2R – S1R = –2 5l
The waves emanating from S1 will arrive exactly two and a half cycles
later than the waves from S2 [Fig |
9 | 1258-1261 | 10 9(b)] for which
S2R – S1R = –2 5l
The waves emanating from S1 will arrive exactly two and a half cycles
later than the waves from S2 [Fig 10 |
9 | 1259-1262 | 9(b)] for which
S2R – S1R = –2 5l
The waves emanating from S1 will arrive exactly two and a half cycles
later than the waves from S2 [Fig 10 10(b)] |
9 | 1260-1263 | 5l
The waves emanating from S1 will arrive exactly two and a half cycles
later than the waves from S2 [Fig 10 10(b)] Thus if the displacement
produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt + 5p) = – a cos wt
FIGURE 10 |
9 | 1261-1264 | 10 10(b)] Thus if the displacement
produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt + 5p) = – a cos wt
FIGURE 10 9
(a) Constructive
interference at a
point Q for which the
path difference is 2l |
9 | 1262-1265 | 10(b)] Thus if the displacement
produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt + 5p) = – a cos wt
FIGURE 10 9
(a) Constructive
interference at a
point Q for which the
path difference is 2l (b) Destructive
interference at a
point R for which the
path difference is
2 |
9 | 1263-1266 | Thus if the displacement
produced by S1 is given by
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt + 5p) = – a cos wt
FIGURE 10 9
(a) Constructive
interference at a
point Q for which the
path difference is 2l (b) Destructive
interference at a
point R for which the
path difference is
2 5 l |
9 | 1264-1267 | 9
(a) Constructive
interference at a
point Q for which the
path difference is 2l (b) Destructive
interference at a
point R for which the
path difference is
2 5 l FIGURE 10 |
9 | 1265-1268 | (b) Destructive
interference at a
point R for which the
path difference is
2 5 l FIGURE 10 10 Locus
of points for which
S1P – S2P is equal to
zero, ±l, ± 2l, ± 3l |
9 | 1266-1269 | 5 l FIGURE 10 10 Locus
of points for which
S1P – S2P is equal to
zero, ±l, ± 2l, ± 3l Rationalised 2023-24
Physics
264
where we have used the fact that a path difference of 2 |
9 | 1267-1270 | FIGURE 10 10 Locus
of points for which
S1P – S2P is equal to
zero, ±l, ± 2l, ± 3l Rationalised 2023-24
Physics
264
where we have used the fact that a path difference of 2 5l corresponds to
a phase difference of 5p |
9 | 1268-1271 | 10 Locus
of points for which
S1P – S2P is equal to
zero, ±l, ± 2l, ± 3l Rationalised 2023-24
Physics
264
where we have used the fact that a path difference of 2 5l corresponds to
a phase difference of 5p The two displacements are now out of phase
and the two displacements will cancel out to give zero intensity |
9 | 1269-1272 | Rationalised 2023-24
Physics
264
where we have used the fact that a path difference of 2 5l corresponds to
a phase difference of 5p The two displacements are now out of phase
and the two displacements will cancel out to give zero intensity This is
referred to as destructive interference |
9 | 1270-1273 | 5l corresponds to
a phase difference of 5p The two displacements are now out of phase
and the two displacements will cancel out to give zero intensity This is
referred to as destructive interference To summarise: If we have two coherent sources S1 and S2 vibrating
in phase, then for an arbitrary point P whenever the path difference,
S1P ~ S2P = nl (n = 0, 1, 2, 3, |
9 | 1271-1274 | The two displacements are now out of phase
and the two displacements will cancel out to give zero intensity This is
referred to as destructive interference To summarise: If we have two coherent sources S1 and S2 vibrating
in phase, then for an arbitrary point P whenever the path difference,
S1P ~ S2P = nl (n = 0, 1, 2, 3, )
(10 |
9 | 1272-1275 | This is
referred to as destructive interference To summarise: If we have two coherent sources S1 and S2 vibrating
in phase, then for an arbitrary point P whenever the path difference,
S1P ~ S2P = nl (n = 0, 1, 2, 3, )
(10 9)
we will have constructive interference and the resultant intensity will be
4I0; the sign ~ between S1P and S2 P represents the difference between
S1P and S2 P |
9 | 1273-1276 | To summarise: If we have two coherent sources S1 and S2 vibrating
in phase, then for an arbitrary point P whenever the path difference,
S1P ~ S2P = nl (n = 0, 1, 2, 3, )
(10 9)
we will have constructive interference and the resultant intensity will be
4I0; the sign ~ between S1P and S2 P represents the difference between
S1P and S2 P On the other hand, if the point P is such that the path
difference,
S1P ~ S2P = (n+ 1
2 ) l (n = 0, 1, 2, 3, |
9 | 1274-1277 | )
(10 9)
we will have constructive interference and the resultant intensity will be
4I0; the sign ~ between S1P and S2 P represents the difference between
S1P and S2 P On the other hand, if the point P is such that the path
difference,
S1P ~ S2P = (n+ 1
2 ) l (n = 0, 1, 2, 3, )
(10 |
9 | 1275-1278 | 9)
we will have constructive interference and the resultant intensity will be
4I0; the sign ~ between S1P and S2 P represents the difference between
S1P and S2 P On the other hand, if the point P is such that the path
difference,
S1P ~ S2P = (n+ 1
2 ) l (n = 0, 1, 2, 3, )
(10 10)
we will have destructive interference and the resultant intensity will be
zero |
9 | 1276-1279 | On the other hand, if the point P is such that the path
difference,
S1P ~ S2P = (n+ 1
2 ) l (n = 0, 1, 2, 3, )
(10 10)
we will have destructive interference and the resultant intensity will be
zero Now, for any other arbitrary point G (Fig |
9 | 1277-1280 | )
(10 10)
we will have destructive interference and the resultant intensity will be
zero Now, for any other arbitrary point G (Fig 10 |
9 | 1278-1281 | 10)
we will have destructive interference and the resultant intensity will be
zero Now, for any other arbitrary point G (Fig 10 10) let the phase
difference between the two displacements be f |
9 | 1279-1282 | Now, for any other arbitrary point G (Fig 10 10) let the phase
difference between the two displacements be f Thus, if the displacement
produced by S1 is given by
y1 = a cos wt
then, the displacement produced by S2 would be
y2 = a cos (wt + f)
and the resultant displacement will be given by
y = y1 + y2
= a [cos wt + cos (wt +f)]
= 2 a cos (f/2) cos (wt + f/2)
The amplitude of the resultant displacement is 2a cos (f/2) and
therefore the intensity at that point will be
I = 4 I0 cos2 (f/2)
(10 |
9 | 1280-1283 | 10 10) let the phase
difference between the two displacements be f Thus, if the displacement
produced by S1 is given by
y1 = a cos wt
then, the displacement produced by S2 would be
y2 = a cos (wt + f)
and the resultant displacement will be given by
y = y1 + y2
= a [cos wt + cos (wt +f)]
= 2 a cos (f/2) cos (wt + f/2)
The amplitude of the resultant displacement is 2a cos (f/2) and
therefore the intensity at that point will be
I = 4 I0 cos2 (f/2)
(10 11)
If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by
Eq |
9 | 1281-1284 | 10) let the phase
difference between the two displacements be f Thus, if the displacement
produced by S1 is given by
y1 = a cos wt
then, the displacement produced by S2 would be
y2 = a cos (wt + f)
and the resultant displacement will be given by
y = y1 + y2
= a [cos wt + cos (wt +f)]
= 2 a cos (f/2) cos (wt + f/2)
The amplitude of the resultant displacement is 2a cos (f/2) and
therefore the intensity at that point will be
I = 4 I0 cos2 (f/2)
(10 11)
If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by
Eq (10 |
9 | 1282-1285 | Thus, if the displacement
produced by S1 is given by
y1 = a cos wt
then, the displacement produced by S2 would be
y2 = a cos (wt + f)
and the resultant displacement will be given by
y = y1 + y2
= a [cos wt + cos (wt +f)]
= 2 a cos (f/2) cos (wt + f/2)
The amplitude of the resultant displacement is 2a cos (f/2) and
therefore the intensity at that point will be
I = 4 I0 cos2 (f/2)
(10 11)
If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by
Eq (10 9) we will have constructive interference leading to maximum
intensity |
9 | 1283-1286 | 11)
If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by
Eq (10 9) we will have constructive interference leading to maximum
intensity On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to
the condition given by Eq |
9 | 1284-1287 | (10 9) we will have constructive interference leading to maximum
intensity On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to
the condition given by Eq (10 |
9 | 1285-1288 | 9) we will have constructive interference leading to maximum
intensity On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to
the condition given by Eq (10 10)] we will have destructive interference
leading to zero intensity |
9 | 1286-1289 | On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to
the condition given by Eq (10 10)] we will have destructive interference
leading to zero intensity Now if the two sources are coherent (i |
9 | 1287-1290 | (10 10)] we will have destructive interference
leading to zero intensity Now if the two sources are coherent (i e |
9 | 1288-1291 | 10)] we will have destructive interference
leading to zero intensity Now if the two sources are coherent (i e , if the two needles are going
up and down regularly) then the phase difference f at any point will not
change with time and we will have a stable interference pattern; i |
9 | 1289-1292 | Now if the two sources are coherent (i e , if the two needles are going
up and down regularly) then the phase difference f at any point will not
change with time and we will have a stable interference pattern; i e |
9 | 1290-1293 | e , if the two needles are going
up and down regularly) then the phase difference f at any point will not
change with time and we will have a stable interference pattern; i e , the
positions of maxima and minima will not change with time |
9 | 1291-1294 | , if the two needles are going
up and down regularly) then the phase difference f at any point will not
change with time and we will have a stable interference pattern; i e , the
positions of maxima and minima will not change with time However, if
the two needles do not maintain a constant phase difference, then the
interference pattern will also change with time and, if the phase difference
changes very rapidly with time, the positions of maxima and minima will
also vary rapidly with time and we will see a “time-averaged” intensity
distribution |
9 | 1292-1295 | e , the
positions of maxima and minima will not change with time However, if
the two needles do not maintain a constant phase difference, then the
interference pattern will also change with time and, if the phase difference
changes very rapidly with time, the positions of maxima and minima will
also vary rapidly with time and we will see a “time-averaged” intensity
distribution When this happens, we will observe an average intensity
that will be given by
I = 2 I0
(10 |
9 | 1293-1296 | , the
positions of maxima and minima will not change with time However, if
the two needles do not maintain a constant phase difference, then the
interference pattern will also change with time and, if the phase difference
changes very rapidly with time, the positions of maxima and minima will
also vary rapidly with time and we will see a “time-averaged” intensity
distribution When this happens, we will observe an average intensity
that will be given by
I = 2 I0
(10 12)
at all points |
9 | 1294-1297 | However, if
the two needles do not maintain a constant phase difference, then the
interference pattern will also change with time and, if the phase difference
changes very rapidly with time, the positions of maxima and minima will
also vary rapidly with time and we will see a “time-averaged” intensity
distribution When this happens, we will observe an average intensity
that will be given by
I = 2 I0
(10 12)
at all points Ripple Tank experiments on wave interference
http://phet |
9 | 1295-1298 | When this happens, we will observe an average intensity
that will be given by
I = 2 I0
(10 12)
at all points Ripple Tank experiments on wave interference
http://phet colorado |
9 | 1296-1299 | 12)
at all points Ripple Tank experiments on wave interference
http://phet colorado edu/en/simulation/legacy/wave-interference
Rationalised 2023-24
265
Wave Optics
When the phase difference between the two vibrating sources changes
rapidly with time, we say that the two sources are incoherent and when
this happens the intensities just add up |
9 | 1297-1300 | Ripple Tank experiments on wave interference
http://phet colorado edu/en/simulation/legacy/wave-interference
Rationalised 2023-24
265
Wave Optics
When the phase difference between the two vibrating sources changes
rapidly with time, we say that the two sources are incoherent and when
this happens the intensities just add up This is indeed what happens
when two separate light sources illuminate a wall |
9 | 1298-1301 | colorado edu/en/simulation/legacy/wave-interference
Rationalised 2023-24
265
Wave Optics
When the phase difference between the two vibrating sources changes
rapidly with time, we say that the two sources are incoherent and when
this happens the intensities just add up This is indeed what happens
when two separate light sources illuminate a wall 10 |
9 | 1299-1302 | edu/en/simulation/legacy/wave-interference
Rationalised 2023-24
265
Wave Optics
When the phase difference between the two vibrating sources changes
rapidly with time, we say that the two sources are incoherent and when
this happens the intensities just add up This is indeed what happens
when two separate light sources illuminate a wall 10 5 INTERFERENCE OF LIGHT WAVES AND YOUNG’S
EXPERIMENT
We will now discuss interference using light waves |
9 | 1300-1303 | This is indeed what happens
when two separate light sources illuminate a wall 10 5 INTERFERENCE OF LIGHT WAVES AND YOUNG’S
EXPERIMENT
We will now discuss interference using light waves If
we use two sodium lamps illuminating two pinholes
(Fig |
9 | 1301-1304 | 10 5 INTERFERENCE OF LIGHT WAVES AND YOUNG’S
EXPERIMENT
We will now discuss interference using light waves If
we use two sodium lamps illuminating two pinholes
(Fig 10 |
9 | 1302-1305 | 5 INTERFERENCE OF LIGHT WAVES AND YOUNG’S
EXPERIMENT
We will now discuss interference using light waves If
we use two sodium lamps illuminating two pinholes
(Fig 10 11) we will not observe any interference fringes |
9 | 1303-1306 | If
we use two sodium lamps illuminating two pinholes
(Fig 10 11) we will not observe any interference fringes This is because of the fact that the light wave emitted
from an ordinary source (like a sodium lamp) undergoes
abrupt phase changes in times of the order of 10–10
seconds |
9 | 1304-1307 | 10 11) we will not observe any interference fringes This is because of the fact that the light wave emitted
from an ordinary source (like a sodium lamp) undergoes
abrupt phase changes in times of the order of 10–10
seconds Thus the light waves coming out from two
independent sources of light will not have any fixed
phase relationship and would be incoherent, when this
happens, as discussed in the previous section, the
intensities on the screen will add up |
9 | 1305-1308 | 11) we will not observe any interference fringes This is because of the fact that the light wave emitted
from an ordinary source (like a sodium lamp) undergoes
abrupt phase changes in times of the order of 10–10
seconds Thus the light waves coming out from two
independent sources of light will not have any fixed
phase relationship and would be incoherent, when this
happens, as discussed in the previous section, the
intensities on the screen will add up The British physicist Thomas Young used an
ingenious technique to “lock” the phases of the waves
emanating from S1 and S2 |
9 | 1306-1309 | This is because of the fact that the light wave emitted
from an ordinary source (like a sodium lamp) undergoes
abrupt phase changes in times of the order of 10–10
seconds Thus the light waves coming out from two
independent sources of light will not have any fixed
phase relationship and would be incoherent, when this
happens, as discussed in the previous section, the
intensities on the screen will add up The British physicist Thomas Young used an
ingenious technique to “lock” the phases of the waves
emanating from S1 and S2 He made two pinholes S1
and S2 (very close to each other) on an opaque screen [Fig |
9 | 1307-1310 | Thus the light waves coming out from two
independent sources of light will not have any fixed
phase relationship and would be incoherent, when this
happens, as discussed in the previous section, the
intensities on the screen will add up The British physicist Thomas Young used an
ingenious technique to “lock” the phases of the waves
emanating from S1 and S2 He made two pinholes S1
and S2 (very close to each other) on an opaque screen [Fig 10 |
9 | 1308-1311 | The British physicist Thomas Young used an
ingenious technique to “lock” the phases of the waves
emanating from S1 and S2 He made two pinholes S1
and S2 (very close to each other) on an opaque screen [Fig 10 12(a)] |
9 | 1309-1312 | He made two pinholes S1
and S2 (very close to each other) on an opaque screen [Fig 10 12(a)] These were illuminated by another pinholes that was in turn, lit by a
bright source |
9 | 1310-1313 | 10 12(a)] These were illuminated by another pinholes that was in turn, lit by a
bright source Light waves spread out from S and fall on both S1 and S2 |
9 | 1311-1314 | 12(a)] These were illuminated by another pinholes that was in turn, lit by a
bright source Light waves spread out from S and fall on both S1 and S2 S1 and S2 then behave like two coherent sources because light waves
coming out from S1 and S2 are derived from the same original source
and any abrupt phase change in S will manifest in exactly similar phase
changes in the light coming out from S1 and S2 |
9 | 1312-1315 | These were illuminated by another pinholes that was in turn, lit by a
bright source Light waves spread out from S and fall on both S1 and S2 S1 and S2 then behave like two coherent sources because light waves
coming out from S1 and S2 are derived from the same original source
and any abrupt phase change in S will manifest in exactly similar phase
changes in the light coming out from S1 and S2 Thus, the two sources S1
and S2 will be locked in phase; i |
9 | 1313-1316 | Light waves spread out from S and fall on both S1 and S2 S1 and S2 then behave like two coherent sources because light waves
coming out from S1 and S2 are derived from the same original source
and any abrupt phase change in S will manifest in exactly similar phase
changes in the light coming out from S1 and S2 Thus, the two sources S1
and S2 will be locked in phase; i e |
9 | 1314-1317 | S1 and S2 then behave like two coherent sources because light waves
coming out from S1 and S2 are derived from the same original source
and any abrupt phase change in S will manifest in exactly similar phase
changes in the light coming out from S1 and S2 Thus, the two sources S1
and S2 will be locked in phase; i e , they will be coherent like the two
vibrating needle in our water wave example [Fig |
9 | 1315-1318 | Thus, the two sources S1
and S2 will be locked in phase; i e , they will be coherent like the two
vibrating needle in our water wave example [Fig 10 |
9 | 1316-1319 | e , they will be coherent like the two
vibrating needle in our water wave example [Fig 10 8(a)] |
9 | 1317-1320 | , they will be coherent like the two
vibrating needle in our water wave example [Fig 10 8(a)] The spherical waves emanating from S1 and S2 will produce
interference fringes on the screen GG¢, as shown in Fig |
9 | 1318-1321 | 10 8(a)] The spherical waves emanating from S1 and S2 will produce
interference fringes on the screen GG¢, as shown in Fig 10 |
9 | 1319-1322 | 8(a)] The spherical waves emanating from S1 and S2 will produce
interference fringes on the screen GG¢, as shown in Fig 10 12(b) |
9 | 1320-1323 | The spherical waves emanating from S1 and S2 will produce
interference fringes on the screen GG¢, as shown in Fig 10 12(b) The
positions of maximum and minimum intensities can be calculated by
using the analysis given in Section 10 |
9 | 1321-1324 | 10 12(b) The
positions of maximum and minimum intensities can be calculated by
using the analysis given in Section 10 4 |
9 | 1322-1325 | 12(b) The
positions of maximum and minimum intensities can be calculated by
using the analysis given in Section 10 4 (a)
(b)
FIGURE 10 |
9 | 1323-1326 | The
positions of maximum and minimum intensities can be calculated by
using the analysis given in Section 10 4 (a)
(b)
FIGURE 10 12 Young’s arrangement to produce interference pattern |
9 | 1324-1327 | 4 (a)
(b)
FIGURE 10 12 Young’s arrangement to produce interference pattern FIGURE 10 |
9 | 1325-1328 | (a)
(b)
FIGURE 10 12 Young’s arrangement to produce interference pattern FIGURE 10 11 If two sodium
lamps illuminate two pinholes
S1 and S2, the intensities will add
up and no interference fringes will
be observed on the screen |
9 | 1326-1329 | 12 Young’s arrangement to produce interference pattern FIGURE 10 11 If two sodium
lamps illuminate two pinholes
S1 and S2, the intensities will add
up and no interference fringes will
be observed on the screen Rationalised 2023-24
Physics
266
FIGURE 10 |
9 | 1327-1330 | FIGURE 10 11 If two sodium
lamps illuminate two pinholes
S1 and S2, the intensities will add
up and no interference fringes will
be observed on the screen Rationalised 2023-24
Physics
266
FIGURE 10 13 Computer generated fringe pattern produced by two point
source S1 and S2 on the screen GG¢ (Fig |
9 | 1328-1331 | 11 If two sodium
lamps illuminate two pinholes
S1 and S2, the intensities will add
up and no interference fringes will
be observed on the screen Rationalised 2023-24
Physics
266
FIGURE 10 13 Computer generated fringe pattern produced by two point
source S1 and S2 on the screen GG¢ (Fig 10 |
9 | 1329-1332 | Rationalised 2023-24
Physics
266
FIGURE 10 13 Computer generated fringe pattern produced by two point
source S1 and S2 on the screen GG¢ (Fig 10 12); correspond to d = 0 |
9 | 1330-1333 | 13 Computer generated fringe pattern produced by two point
source S1 and S2 on the screen GG¢ (Fig 10 12); correspond to d = 0 025
mm, D = 5 cm and l = 5 × 10–5 cm |
9 | 1331-1334 | 10 12); correspond to d = 0 025
mm, D = 5 cm and l = 5 × 10–5 cm ) (Adopted from OPTICS by A |
9 | 1332-1335 | 12); correspond to d = 0 025
mm, D = 5 cm and l = 5 × 10–5 cm ) (Adopted from OPTICS by A Ghatak,
Tata McGraw Hill Publishing Co |
9 | 1333-1336 | 025
mm, D = 5 cm and l = 5 × 10–5 cm ) (Adopted from OPTICS by A Ghatak,
Tata McGraw Hill Publishing Co Ltd |
9 | 1334-1337 | ) (Adopted from OPTICS by A Ghatak,
Tata McGraw Hill Publishing Co Ltd , New Delhi, 2000 |
9 | 1335-1338 | Ghatak,
Tata McGraw Hill Publishing Co Ltd , New Delhi, 2000 )
Thomas
Young
(1773 – 1829) English
physicist, physician and
Egyptologist |
9 | 1336-1339 | Ltd , New Delhi, 2000 )
Thomas
Young
(1773 – 1829) English
physicist, physician and
Egyptologist Young worked
on a wide variety of
scientific problems, ranging
from the structure of the eye
and the mechanism of
vision to the decipherment
of the Rosetta stone |
9 | 1337-1340 | , New Delhi, 2000 )
Thomas
Young
(1773 – 1829) English
physicist, physician and
Egyptologist Young worked
on a wide variety of
scientific problems, ranging
from the structure of the eye
and the mechanism of
vision to the decipherment
of the Rosetta stone He
revived the wave theory of
light and recognised that
interference phenomena
provide proof of the wave
properties of light |
9 | 1338-1341 | )
Thomas
Young
(1773 – 1829) English
physicist, physician and
Egyptologist Young worked
on a wide variety of
scientific problems, ranging
from the structure of the eye
and the mechanism of
vision to the decipherment
of the Rosetta stone He
revived the wave theory of
light and recognised that
interference phenomena
provide proof of the wave
properties of light THOMAS YOUNG (1773 – 1829)
We will have constructive interference resulting in a bright
region when xd
D
= nl |
Subsets and Splits