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9 | 439-442 | Note that the
formula is true for a concave lens also In
that case R1is negative, R 2 positive and
therefore, f is negative FIGURE 9 16 (a) The position of object, and the
image formed by a double convex lens,
(b) Refraction at the first spherical surface and
(c) Refraction at the second spherical surface |
9 | 440-443 | In
that case R1is negative, R 2 positive and
therefore, f is negative FIGURE 9 16 (a) The position of object, and the
image formed by a double convex lens,
(b) Refraction at the first spherical surface and
(c) Refraction at the second spherical surface *
Note that now the refractive index of the medium on the right side of ADC is n1
while on its left it is n2 |
9 | 441-444 | FIGURE 9 16 (a) The position of object, and the
image formed by a double convex lens,
(b) Refraction at the first spherical surface and
(c) Refraction at the second spherical surface *
Note that now the refractive index of the medium on the right side of ADC is n1
while on its left it is n2 Further DI1 is negative as the distance is measured
against the direction of incident light |
9 | 442-445 | 16 (a) The position of object, and the
image formed by a double convex lens,
(b) Refraction at the first spherical surface and
(c) Refraction at the second spherical surface *
Note that now the refractive index of the medium on the right side of ADC is n1
while on its left it is n2 Further DI1 is negative as the distance is measured
against the direction of incident light Rationalised 2023-24
Ray Optics and
Optical Instruments
235
From Eqs |
9 | 443-446 | *
Note that now the refractive index of the medium on the right side of ADC is n1
while on its left it is n2 Further DI1 is negative as the distance is measured
against the direction of incident light Rationalised 2023-24
Ray Optics and
Optical Instruments
235
From Eqs (9 |
9 | 444-447 | Further DI1 is negative as the distance is measured
against the direction of incident light Rationalised 2023-24
Ray Optics and
Optical Instruments
235
From Eqs (9 19) and (9 |
9 | 445-448 | Rationalised 2023-24
Ray Optics and
Optical Instruments
235
From Eqs (9 19) and (9 20), we get
1
1
1
OB
DI
n
n
fn
+
=
(9 |
9 | 446-449 | (9 19) and (9 20), we get
1
1
1
OB
DI
n
n
fn
+
=
(9 22)
Again, in the thin lens approximation, B and D are both close to the
optical centre of the lens |
9 | 447-450 | 19) and (9 20), we get
1
1
1
OB
DI
n
n
fn
+
=
(9 22)
Again, in the thin lens approximation, B and D are both close to the
optical centre of the lens Applying the sign convention,
BO = – u, DI = +v, we get
1
1
1
v
u
f
−
=
(9 |
9 | 448-451 | 20), we get
1
1
1
OB
DI
n
n
fn
+
=
(9 22)
Again, in the thin lens approximation, B and D are both close to the
optical centre of the lens Applying the sign convention,
BO = – u, DI = +v, we get
1
1
1
v
u
f
−
=
(9 23)
Equation (9 |
9 | 449-452 | 22)
Again, in the thin lens approximation, B and D are both close to the
optical centre of the lens Applying the sign convention,
BO = – u, DI = +v, we get
1
1
1
v
u
f
−
=
(9 23)
Equation (9 23) is the familiar thin lens formula |
9 | 450-453 | Applying the sign convention,
BO = – u, DI = +v, we get
1
1
1
v
u
f
−
=
(9 23)
Equation (9 23) is the familiar thin lens formula Though we derived
it for a real image formed by a convex lens, the formula is valid for both
convex as well as concave lenses and for both real and virtual images |
9 | 451-454 | 23)
Equation (9 23) is the familiar thin lens formula Though we derived
it for a real image formed by a convex lens, the formula is valid for both
convex as well as concave lenses and for both real and virtual images It is worth mentioning that the two foci, F and F¢, of a double convex
or concave lens are equidistant from the optical centre |
9 | 452-455 | 23) is the familiar thin lens formula Though we derived
it for a real image formed by a convex lens, the formula is valid for both
convex as well as concave lenses and for both real and virtual images It is worth mentioning that the two foci, F and F¢, of a double convex
or concave lens are equidistant from the optical centre The focus on the
side of the (original) source of light is called the first focal point, whereas
the other is called the second focal point |
9 | 453-456 | Though we derived
it for a real image formed by a convex lens, the formula is valid for both
convex as well as concave lenses and for both real and virtual images It is worth mentioning that the two foci, F and F¢, of a double convex
or concave lens are equidistant from the optical centre The focus on the
side of the (original) source of light is called the first focal point, whereas
the other is called the second focal point To find the image of an object by a lens, we can, in principle, take any
two rays emanating from a point on an object; trace their paths using the
laws of refraction and find the point where the refracted rays meet (or
appear to meet) |
9 | 454-457 | It is worth mentioning that the two foci, F and F¢, of a double convex
or concave lens are equidistant from the optical centre The focus on the
side of the (original) source of light is called the first focal point, whereas
the other is called the second focal point To find the image of an object by a lens, we can, in principle, take any
two rays emanating from a point on an object; trace their paths using the
laws of refraction and find the point where the refracted rays meet (or
appear to meet) In practice, however, it is convenient to choose any two
of the following rays:
(i)
A ray emanating from the object parallel to the principal axis of the
lens after refraction passes through the second principal focus F¢ (in
a convex lens) or appears to diverge (in a concave lens) from the first
principal focus F |
9 | 455-458 | The focus on the
side of the (original) source of light is called the first focal point, whereas
the other is called the second focal point To find the image of an object by a lens, we can, in principle, take any
two rays emanating from a point on an object; trace their paths using the
laws of refraction and find the point where the refracted rays meet (or
appear to meet) In practice, however, it is convenient to choose any two
of the following rays:
(i)
A ray emanating from the object parallel to the principal axis of the
lens after refraction passes through the second principal focus F¢ (in
a convex lens) or appears to diverge (in a concave lens) from the first
principal focus F (ii) A ray of light, passing through the optical
centre of the lens, emerges without any
deviation after refraction |
9 | 456-459 | To find the image of an object by a lens, we can, in principle, take any
two rays emanating from a point on an object; trace their paths using the
laws of refraction and find the point where the refracted rays meet (or
appear to meet) In practice, however, it is convenient to choose any two
of the following rays:
(i)
A ray emanating from the object parallel to the principal axis of the
lens after refraction passes through the second principal focus F¢ (in
a convex lens) or appears to diverge (in a concave lens) from the first
principal focus F (ii) A ray of light, passing through the optical
centre of the lens, emerges without any
deviation after refraction (iii) (a) A ray of light passing through the first
principal focus of a convex lens [Fig |
9 | 457-460 | In practice, however, it is convenient to choose any two
of the following rays:
(i)
A ray emanating from the object parallel to the principal axis of the
lens after refraction passes through the second principal focus F¢ (in
a convex lens) or appears to diverge (in a concave lens) from the first
principal focus F (ii) A ray of light, passing through the optical
centre of the lens, emerges without any
deviation after refraction (iii) (a) A ray of light passing through the first
principal focus of a convex lens [Fig 9 |
9 | 458-461 | (ii) A ray of light, passing through the optical
centre of the lens, emerges without any
deviation after refraction (iii) (a) A ray of light passing through the first
principal focus of a convex lens [Fig 9 17(a)]
emerges parallel to the principal axis after
refraction |
9 | 459-462 | (iii) (a) A ray of light passing through the first
principal focus of a convex lens [Fig 9 17(a)]
emerges parallel to the principal axis after
refraction (b) A ray of light incident on a concave lens
appearing to meet the principal axis at
second focus point emerges parallel to the
principal axis after refraction [Fig |
9 | 460-463 | 9 17(a)]
emerges parallel to the principal axis after
refraction (b) A ray of light incident on a concave lens
appearing to meet the principal axis at
second focus point emerges parallel to the
principal axis after refraction [Fig 9 |
9 | 461-464 | 17(a)]
emerges parallel to the principal axis after
refraction (b) A ray of light incident on a concave lens
appearing to meet the principal axis at
second focus point emerges parallel to the
principal axis after refraction [Fig 9 17(b)] |
9 | 462-465 | (b) A ray of light incident on a concave lens
appearing to meet the principal axis at
second focus point emerges parallel to the
principal axis after refraction [Fig 9 17(b)] Figures 9 |
9 | 463-466 | 9 17(b)] Figures 9 17(a) and (b) illustrate these rules
for a convex and a concave lens, respectively |
9 | 464-467 | 17(b)] Figures 9 17(a) and (b) illustrate these rules
for a convex and a concave lens, respectively You should practice drawing similar ray
diagrams for different positions of the object with
respect to the lens and also verify that the lens
formula, Eq |
9 | 465-468 | Figures 9 17(a) and (b) illustrate these rules
for a convex and a concave lens, respectively You should practice drawing similar ray
diagrams for different positions of the object with
respect to the lens and also verify that the lens
formula, Eq (9 |
9 | 466-469 | 17(a) and (b) illustrate these rules
for a convex and a concave lens, respectively You should practice drawing similar ray
diagrams for different positions of the object with
respect to the lens and also verify that the lens
formula, Eq (9 23), holds good for all cases |
9 | 467-470 | You should practice drawing similar ray
diagrams for different positions of the object with
respect to the lens and also verify that the lens
formula, Eq (9 23), holds good for all cases Here again it must be remembered that each
point on an object gives out infinite number of
rays |
9 | 468-471 | (9 23), holds good for all cases Here again it must be remembered that each
point on an object gives out infinite number of
rays All these rays will pass through the same
image point after refraction at the lens |
9 | 469-472 | 23), holds good for all cases Here again it must be remembered that each
point on an object gives out infinite number of
rays All these rays will pass through the same
image point after refraction at the lens Magnification (m) produced by a lens is
defined, like that for a mirror, as the ratio of the
size of the image to that of the object |
9 | 470-473 | Here again it must be remembered that each
point on an object gives out infinite number of
rays All these rays will pass through the same
image point after refraction at the lens Magnification (m) produced by a lens is
defined, like that for a mirror, as the ratio of the
size of the image to that of the object Proceeding
FIGURE 9 |
9 | 471-474 | All these rays will pass through the same
image point after refraction at the lens Magnification (m) produced by a lens is
defined, like that for a mirror, as the ratio of the
size of the image to that of the object Proceeding
FIGURE 9 17 Tracing rays through (a)
convex lens (b) concave lens |
9 | 472-475 | Magnification (m) produced by a lens is
defined, like that for a mirror, as the ratio of the
size of the image to that of the object Proceeding
FIGURE 9 17 Tracing rays through (a)
convex lens (b) concave lens Rationalised 2023-24
Physics
236
EXAMPLE 9 |
9 | 473-476 | Proceeding
FIGURE 9 17 Tracing rays through (a)
convex lens (b) concave lens Rationalised 2023-24
Physics
236
EXAMPLE 9 6
in the same way as for spherical mirrors, it is easily seen that
for a lens
m =
h
h
′ =
uv
(9 |
9 | 474-477 | 17 Tracing rays through (a)
convex lens (b) concave lens Rationalised 2023-24
Physics
236
EXAMPLE 9 6
in the same way as for spherical mirrors, it is easily seen that
for a lens
m =
h
h
′ =
uv
(9 24)
When we apply the sign convention, we see that, for erect (and virtual)
image formed by a convex or concave lens, m is positive, while for an
inverted (and real) image, m is negative |
9 | 475-478 | Rationalised 2023-24
Physics
236
EXAMPLE 9 6
in the same way as for spherical mirrors, it is easily seen that
for a lens
m =
h
h
′ =
uv
(9 24)
When we apply the sign convention, we see that, for erect (and virtual)
image formed by a convex or concave lens, m is positive, while for an
inverted (and real) image, m is negative Example 9 |
9 | 476-479 | 6
in the same way as for spherical mirrors, it is easily seen that
for a lens
m =
h
h
′ =
uv
(9 24)
When we apply the sign convention, we see that, for erect (and virtual)
image formed by a convex or concave lens, m is positive, while for an
inverted (and real) image, m is negative Example 9 6 A magician during a show makes a glass lens with
n = 1 |
9 | 477-480 | 24)
When we apply the sign convention, we see that, for erect (and virtual)
image formed by a convex or concave lens, m is positive, while for an
inverted (and real) image, m is negative Example 9 6 A magician during a show makes a glass lens with
n = 1 47 disappear in a trough of liquid |
9 | 478-481 | Example 9 6 A magician during a show makes a glass lens with
n = 1 47 disappear in a trough of liquid What is the refractive index
of the liquid |
9 | 479-482 | 6 A magician during a show makes a glass lens with
n = 1 47 disappear in a trough of liquid What is the refractive index
of the liquid Could the liquid be water |
9 | 480-483 | 47 disappear in a trough of liquid What is the refractive index
of the liquid Could the liquid be water Solution
The refractive index of the liquid must be equal to 1 |
9 | 481-484 | What is the refractive index
of the liquid Could the liquid be water Solution
The refractive index of the liquid must be equal to 1 47 in order to
make the lens disappear |
9 | 482-485 | Could the liquid be water Solution
The refractive index of the liquid must be equal to 1 47 in order to
make the lens disappear This means n1 = n2 |
9 | 483-486 | Solution
The refractive index of the liquid must be equal to 1 47 in order to
make the lens disappear This means n1 = n2 This gives 1/f =0 or
f ® ¥ |
9 | 484-487 | 47 in order to
make the lens disappear This means n1 = n2 This gives 1/f =0 or
f ® ¥ The lens in the liquid will act like a plane sheet of glass |
9 | 485-488 | This means n1 = n2 This gives 1/f =0 or
f ® ¥ The lens in the liquid will act like a plane sheet of glass No,
the liquid is not water |
9 | 486-489 | This gives 1/f =0 or
f ® ¥ The lens in the liquid will act like a plane sheet of glass No,
the liquid is not water It could be glycerine |
9 | 487-490 | The lens in the liquid will act like a plane sheet of glass No,
the liquid is not water It could be glycerine FIGURE 9 |
9 | 488-491 | No,
the liquid is not water It could be glycerine FIGURE 9 18 Power of a lens |
9 | 489-492 | It could be glycerine FIGURE 9 18 Power of a lens 9 |
9 | 490-493 | FIGURE 9 18 Power of a lens 9 5 |
9 | 491-494 | 18 Power of a lens 9 5 3 Power of a lens
Power of a lens is a measure of the convergence or
divergence, which a lens introduces in the light falling on
it |
9 | 492-495 | 9 5 3 Power of a lens
Power of a lens is a measure of the convergence or
divergence, which a lens introduces in the light falling on
it Clearly, a lens of shorter focal length bends the incident
light more, while converging it in case of a convex lens
and diverging it in case of a concave lens |
9 | 493-496 | 5 3 Power of a lens
Power of a lens is a measure of the convergence or
divergence, which a lens introduces in the light falling on
it Clearly, a lens of shorter focal length bends the incident
light more, while converging it in case of a convex lens
and diverging it in case of a concave lens The power P of
a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light parallel to the
principal axis falling at unit distance from the optical
centre (Fig |
9 | 494-497 | 3 Power of a lens
Power of a lens is a measure of the convergence or
divergence, which a lens introduces in the light falling on
it Clearly, a lens of shorter focal length bends the incident
light more, while converging it in case of a convex lens
and diverging it in case of a concave lens The power P of
a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light parallel to the
principal axis falling at unit distance from the optical
centre (Fig 9 |
9 | 495-498 | Clearly, a lens of shorter focal length bends the incident
light more, while converging it in case of a convex lens
and diverging it in case of a concave lens The power P of
a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light parallel to the
principal axis falling at unit distance from the optical
centre (Fig 9 18) |
9 | 496-499 | The power P of
a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light parallel to the
principal axis falling at unit distance from the optical
centre (Fig 9 18) tan
;
,
tan
δ
δ
=
=
=
fh
h
f
if
1
1
or
δ =f1
for small
value of d |
9 | 497-500 | 9 18) tan
;
,
tan
δ
δ
=
=
=
fh
h
f
if
1
1
or
δ =f1
for small
value of d Thus,
P =
f1
(9 |
9 | 498-501 | 18) tan
;
,
tan
δ
δ
=
=
=
fh
h
f
if
1
1
or
δ =f1
for small
value of d Thus,
P =
f1
(9 25)
The SI unit for power of a lens is dioptre (D): 1D = 1m–1 |
9 | 499-502 | tan
;
,
tan
δ
δ
=
=
=
fh
h
f
if
1
1
or
δ =f1
for small
value of d Thus,
P =
f1
(9 25)
The SI unit for power of a lens is dioptre (D): 1D = 1m–1 The power of
a lens of focal length of 1 metre is one dioptre |
9 | 500-503 | Thus,
P =
f1
(9 25)
The SI unit for power of a lens is dioptre (D): 1D = 1m–1 The power of
a lens of focal length of 1 metre is one dioptre Power of a lens is positive
for a converging lens and negative for a diverging lens |
9 | 501-504 | 25)
The SI unit for power of a lens is dioptre (D): 1D = 1m–1 The power of
a lens of focal length of 1 metre is one dioptre Power of a lens is positive
for a converging lens and negative for a diverging lens Thus, when an
optician prescribes a corrective lens of power + 2 |
9 | 502-505 | The power of
a lens of focal length of 1 metre is one dioptre Power of a lens is positive
for a converging lens and negative for a diverging lens Thus, when an
optician prescribes a corrective lens of power + 2 5 D, the required lens is
a convex lens of focal length + 40 cm |
9 | 503-506 | Power of a lens is positive
for a converging lens and negative for a diverging lens Thus, when an
optician prescribes a corrective lens of power + 2 5 D, the required lens is
a convex lens of focal length + 40 cm A lens of power of – 4 |
9 | 504-507 | Thus, when an
optician prescribes a corrective lens of power + 2 5 D, the required lens is
a convex lens of focal length + 40 cm A lens of power of – 4 0 D means a
concave lens of focal length – 25 cm |
9 | 505-508 | 5 D, the required lens is
a convex lens of focal length + 40 cm A lens of power of – 4 0 D means a
concave lens of focal length – 25 cm Example 9 |
9 | 506-509 | A lens of power of – 4 0 D means a
concave lens of focal length – 25 cm Example 9 7 (i) If f = 0 |
9 | 507-510 | 0 D means a
concave lens of focal length – 25 cm Example 9 7 (i) If f = 0 5 m for a glass lens, what is the power of the
lens |
9 | 508-511 | Example 9 7 (i) If f = 0 5 m for a glass lens, what is the power of the
lens (ii) The radii of curvature of the faces of a double convex lens
are 10 cm and 15 cm |
9 | 509-512 | 7 (i) If f = 0 5 m for a glass lens, what is the power of the
lens (ii) The radii of curvature of the faces of a double convex lens
are 10 cm and 15 cm Its focal length is 12 cm |
9 | 510-513 | 5 m for a glass lens, what is the power of the
lens (ii) The radii of curvature of the faces of a double convex lens
are 10 cm and 15 cm Its focal length is 12 cm What is the refractive
index of glass |
9 | 511-514 | (ii) The radii of curvature of the faces of a double convex lens
are 10 cm and 15 cm Its focal length is 12 cm What is the refractive
index of glass (iii) A convex lens has 20 cm focal length in air |
9 | 512-515 | Its focal length is 12 cm What is the refractive
index of glass (iii) A convex lens has 20 cm focal length in air What
is focal length in water |
9 | 513-516 | What is the refractive
index of glass (iii) A convex lens has 20 cm focal length in air What
is focal length in water (Refractive index of air-water = 1 |
9 | 514-517 | (iii) A convex lens has 20 cm focal length in air What
is focal length in water (Refractive index of air-water = 1 33, refractive
index for air-glass = 1 |
9 | 515-518 | What
is focal length in water (Refractive index of air-water = 1 33, refractive
index for air-glass = 1 5 |
9 | 516-519 | (Refractive index of air-water = 1 33, refractive
index for air-glass = 1 5 )
EXAMPLE 9 |
9 | 517-520 | 33, refractive
index for air-glass = 1 5 )
EXAMPLE 9 7
Rationalised 2023-24
Ray Optics and
Optical Instruments
237
Solution
(i)
Power = +2 dioptre |
9 | 518-521 | 5 )
EXAMPLE 9 7
Rationalised 2023-24
Ray Optics and
Optical Instruments
237
Solution
(i)
Power = +2 dioptre (ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm |
9 | 519-522 | )
EXAMPLE 9 7
Rationalised 2023-24
Ray Optics and
Optical Instruments
237
Solution
(i)
Power = +2 dioptre (ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm Refractive index of air is taken as unity |
9 | 520-523 | 7
Rationalised 2023-24
Ray Optics and
Optical Instruments
237
Solution
(i)
Power = +2 dioptre (ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm Refractive index of air is taken as unity We use the lens formula of Eq |
9 | 521-524 | (ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm Refractive index of air is taken as unity We use the lens formula of Eq (9 |
9 | 522-525 | Refractive index of air is taken as unity We use the lens formula of Eq (9 22) |
9 | 523-526 | We use the lens formula of Eq (9 22) The sign convention has to
be applied for f, R1 and R2 |
9 | 524-527 | (9 22) The sign convention has to
be applied for f, R1 and R2 Substituting the values, we have
1
12
1
1
10
1
15
=
−
− −
(
)
n
This gives n = 1 |
9 | 525-528 | 22) The sign convention has to
be applied for f, R1 and R2 Substituting the values, we have
1
12
1
1
10
1
15
=
−
− −
(
)
n
This gives n = 1 5 |
9 | 526-529 | The sign convention has to
be applied for f, R1 and R2 Substituting the values, we have
1
12
1
1
10
1
15
=
−
− −
(
)
n
This gives n = 1 5 (iii) For a glass lens in air, n2 = 1 |
9 | 527-530 | Substituting the values, we have
1
12
1
1
10
1
15
=
−
− −
(
)
n
This gives n = 1 5 (iii) For a glass lens in air, n2 = 1 5, n1 = 1, f = +20 cm |
9 | 528-531 | 5 (iii) For a glass lens in air, n2 = 1 5, n1 = 1, f = +20 cm Hence, the lens
formula gives
1
20
0 5
1
1
1
2
=
−
|
9 | 529-532 | (iii) For a glass lens in air, n2 = 1 5, n1 = 1, f = +20 cm Hence, the lens
formula gives
1
20
0 5
1
1
1
2
=
−
R
R
For the same glass lens in water, n2 = 1 |
9 | 530-533 | 5, n1 = 1, f = +20 cm Hence, the lens
formula gives
1
20
0 5
1
1
1
2
=
−
R
R
For the same glass lens in water, n2 = 1 5, n1 = 1 |
9 | 531-534 | Hence, the lens
formula gives
1
20
0 5
1
1
1
2
=
−
R
R
For the same glass lens in water, n2 = 1 5, n1 = 1 33 |
9 | 532-535 | R
R
For the same glass lens in water, n2 = 1 5, n1 = 1 33 Therefore,
1 33
1 5 1 33
1
1
1
2 |
9 | 533-536 | 5, n1 = 1 33 Therefore,
1 33
1 5 1 33
1
1
1
2 ( |
9 | 534-537 | 33 Therefore,
1 33
1 5 1 33
1
1
1
2 ( )
f
R
R
=
−
−
(9 |
9 | 535-538 | Therefore,
1 33
1 5 1 33
1
1
1
2 ( )
f
R
R
=
−
−
(9 26)
Combining these two equations, we find f = + 78 |
9 | 536-539 | ( )
f
R
R
=
−
−
(9 26)
Combining these two equations, we find f = + 78 2 cm |
9 | 537-540 | )
f
R
R
=
−
−
(9 26)
Combining these two equations, we find f = + 78 2 cm 9 |
9 | 538-541 | 26)
Combining these two equations, we find f = + 78 2 cm 9 5 |
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