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9 | 339-342 | A bundle of optical fibres can be put to
several uses Optical fibres are extensively
used for transmitting and receiving
FIGURE 9 12
Observing total internal
reflection in water with
a laser beam (refraction
due to glass of beaker
neglected being very
thin) FIGURE 9 |
9 | 340-343 | Optical fibres are extensively
used for transmitting and receiving
FIGURE 9 12
Observing total internal
reflection in water with
a laser beam (refraction
due to glass of beaker
neglected being very
thin) FIGURE 9 13 Prisms designed to bend rays by
90° and 180° or to invert image without changing
its size make use of total internal reflection |
9 | 341-344 | 12
Observing total internal
reflection in water with
a laser beam (refraction
due to glass of beaker
neglected being very
thin) FIGURE 9 13 Prisms designed to bend rays by
90° and 180° or to invert image without changing
its size make use of total internal reflection Rationalised 2023-24
Physics
232
electrical signals which are converted to light
by suitable transducers |
9 | 342-345 | FIGURE 9 13 Prisms designed to bend rays by
90° and 180° or to invert image without changing
its size make use of total internal reflection Rationalised 2023-24
Physics
232
electrical signals which are converted to light
by suitable transducers Obviously, optical
fibres can also be used for transmission of
optical signals |
9 | 343-346 | 13 Prisms designed to bend rays by
90° and 180° or to invert image without changing
its size make use of total internal reflection Rationalised 2023-24
Physics
232
electrical signals which are converted to light
by suitable transducers Obviously, optical
fibres can also be used for transmission of
optical signals For example, these are used
as a ‘light pipe’ to facilitate visual examination
of internal organs like esophagus, stomach
and intestines |
9 | 344-347 | Rationalised 2023-24
Physics
232
electrical signals which are converted to light
by suitable transducers Obviously, optical
fibres can also be used for transmission of
optical signals For example, these are used
as a ‘light pipe’ to facilitate visual examination
of internal organs like esophagus, stomach
and intestines You might have seen a
commonly available decorative lamp with fine
plastic fibres with their free ends forming a
fountain like structure |
9 | 345-348 | Obviously, optical
fibres can also be used for transmission of
optical signals For example, these are used
as a ‘light pipe’ to facilitate visual examination
of internal organs like esophagus, stomach
and intestines You might have seen a
commonly available decorative lamp with fine
plastic fibres with their free ends forming a
fountain like structure The other end of the
fibres is fixed over an electric lamp |
9 | 346-349 | For example, these are used
as a ‘light pipe’ to facilitate visual examination
of internal organs like esophagus, stomach
and intestines You might have seen a
commonly available decorative lamp with fine
plastic fibres with their free ends forming a
fountain like structure The other end of the
fibres is fixed over an electric lamp When the
lamp is switched on, the light travels from the bottom of each fibre and
appears at the tip of its free end as a dot of light |
9 | 347-350 | You might have seen a
commonly available decorative lamp with fine
plastic fibres with their free ends forming a
fountain like structure The other end of the
fibres is fixed over an electric lamp When the
lamp is switched on, the light travels from the bottom of each fibre and
appears at the tip of its free end as a dot of light The fibres in such
decorative lamps are optical fibres |
9 | 348-351 | The other end of the
fibres is fixed over an electric lamp When the
lamp is switched on, the light travels from the bottom of each fibre and
appears at the tip of its free end as a dot of light The fibres in such
decorative lamps are optical fibres The main requirement in fabricating optical fibres is that there should
be very little absorption of light as it travels for long distances inside
them |
9 | 349-352 | When the
lamp is switched on, the light travels from the bottom of each fibre and
appears at the tip of its free end as a dot of light The fibres in such
decorative lamps are optical fibres The main requirement in fabricating optical fibres is that there should
be very little absorption of light as it travels for long distances inside
them This has been achieved by purification and special preparation of
materials such as quartz |
9 | 350-353 | The fibres in such
decorative lamps are optical fibres The main requirement in fabricating optical fibres is that there should
be very little absorption of light as it travels for long distances inside
them This has been achieved by purification and special preparation of
materials such as quartz In silica glass fibres, it is possible to transmit
more than 95% of the light over a fibre length of 1 km |
9 | 351-354 | The main requirement in fabricating optical fibres is that there should
be very little absorption of light as it travels for long distances inside
them This has been achieved by purification and special preparation of
materials such as quartz In silica glass fibres, it is possible to transmit
more than 95% of the light over a fibre length of 1 km (Compare with
what you expect for a block of ordinary window glass 1 km thick |
9 | 352-355 | This has been achieved by purification and special preparation of
materials such as quartz In silica glass fibres, it is possible to transmit
more than 95% of the light over a fibre length of 1 km (Compare with
what you expect for a block of ordinary window glass 1 km thick )
9 |
9 | 353-356 | In silica glass fibres, it is possible to transmit
more than 95% of the light over a fibre length of 1 km (Compare with
what you expect for a block of ordinary window glass 1 km thick )
9 5 REFRACTION AT SPHERICAL SURFACES
AND BY LENSES
We have so far considered refraction at a plane interface |
9 | 354-357 | (Compare with
what you expect for a block of ordinary window glass 1 km thick )
9 5 REFRACTION AT SPHERICAL SURFACES
AND BY LENSES
We have so far considered refraction at a plane interface We shall now
consider refraction at a spherical interface between two transparent media |
9 | 355-358 | )
9 5 REFRACTION AT SPHERICAL SURFACES
AND BY LENSES
We have so far considered refraction at a plane interface We shall now
consider refraction at a spherical interface between two transparent media An infinitesimal part of a spherical surface can be regarded as planar
and the same laws of refraction can be applied at every point on the
surface |
9 | 356-359 | 5 REFRACTION AT SPHERICAL SURFACES
AND BY LENSES
We have so far considered refraction at a plane interface We shall now
consider refraction at a spherical interface between two transparent media An infinitesimal part of a spherical surface can be regarded as planar
and the same laws of refraction can be applied at every point on the
surface Just as for reflection by a spherical mirror, the normal at the
point of incidence is perpendicular to the tangent plane to the spherical
surface at that point and, therefore, passes through its centre of
curvature |
9 | 357-360 | We shall now
consider refraction at a spherical interface between two transparent media An infinitesimal part of a spherical surface can be regarded as planar
and the same laws of refraction can be applied at every point on the
surface Just as for reflection by a spherical mirror, the normal at the
point of incidence is perpendicular to the tangent plane to the spherical
surface at that point and, therefore, passes through its centre of
curvature We first consider refraction by a single spherical surface and
follow it by thin lenses |
9 | 358-361 | An infinitesimal part of a spherical surface can be regarded as planar
and the same laws of refraction can be applied at every point on the
surface Just as for reflection by a spherical mirror, the normal at the
point of incidence is perpendicular to the tangent plane to the spherical
surface at that point and, therefore, passes through its centre of
curvature We first consider refraction by a single spherical surface and
follow it by thin lenses A thin lens is a transparent optical medium
bounded by two surfaces; at least one of which should be spherical |
9 | 359-362 | Just as for reflection by a spherical mirror, the normal at the
point of incidence is perpendicular to the tangent plane to the spherical
surface at that point and, therefore, passes through its centre of
curvature We first consider refraction by a single spherical surface and
follow it by thin lenses A thin lens is a transparent optical medium
bounded by two surfaces; at least one of which should be spherical Applying the formula for image formation by a single spherical surface
successively at the two surfaces of a lens, we shall obtain the lens maker’s
formula and then the lens formula |
9 | 360-363 | We first consider refraction by a single spherical surface and
follow it by thin lenses A thin lens is a transparent optical medium
bounded by two surfaces; at least one of which should be spherical Applying the formula for image formation by a single spherical surface
successively at the two surfaces of a lens, we shall obtain the lens maker’s
formula and then the lens formula 9 |
9 | 361-364 | A thin lens is a transparent optical medium
bounded by two surfaces; at least one of which should be spherical Applying the formula for image formation by a single spherical surface
successively at the two surfaces of a lens, we shall obtain the lens maker’s
formula and then the lens formula 9 5 |
9 | 362-365 | Applying the formula for image formation by a single spherical surface
successively at the two surfaces of a lens, we shall obtain the lens maker’s
formula and then the lens formula 9 5 1 Refraction at a spherical surface
Figure 9 |
9 | 363-366 | 9 5 1 Refraction at a spherical surface
Figure 9 15 shows the geometry of formation of image I of an object O on
the principal axis of a spherical surface with centre of curvature C, and
radius of curvature R |
9 | 364-367 | 5 1 Refraction at a spherical surface
Figure 9 15 shows the geometry of formation of image I of an object O on
the principal axis of a spherical surface with centre of curvature C, and
radius of curvature R The rays are incident from a medium of refractive
index n1, to another of refractive index n 2 |
9 | 365-368 | 1 Refraction at a spherical surface
Figure 9 15 shows the geometry of formation of image I of an object O on
the principal axis of a spherical surface with centre of curvature C, and
radius of curvature R The rays are incident from a medium of refractive
index n1, to another of refractive index n 2 As before, we take the aperture
(or the lateral size) of the surface to be small compared to other distances
involved, so that small angle approximation can be made |
9 | 366-369 | 15 shows the geometry of formation of image I of an object O on
the principal axis of a spherical surface with centre of curvature C, and
radius of curvature R The rays are incident from a medium of refractive
index n1, to another of refractive index n 2 As before, we take the aperture
(or the lateral size) of the surface to be small compared to other distances
involved, so that small angle approximation can be made In particular,
NM will be taken to be nearly equal to the length of the perpendicular
from the point N on the principal axis |
9 | 367-370 | The rays are incident from a medium of refractive
index n1, to another of refractive index n 2 As before, we take the aperture
(or the lateral size) of the surface to be small compared to other distances
involved, so that small angle approximation can be made In particular,
NM will be taken to be nearly equal to the length of the perpendicular
from the point N on the principal axis We have, for small angles,
tan ÐNOM = MN
OM
FIGURE 9 |
9 | 368-371 | As before, we take the aperture
(or the lateral size) of the surface to be small compared to other distances
involved, so that small angle approximation can be made In particular,
NM will be taken to be nearly equal to the length of the perpendicular
from the point N on the principal axis We have, for small angles,
tan ÐNOM = MN
OM
FIGURE 9 14 Light undergoes successive total
internal reflections as it moves through an
optical fibre |
9 | 369-372 | In particular,
NM will be taken to be nearly equal to the length of the perpendicular
from the point N on the principal axis We have, for small angles,
tan ÐNOM = MN
OM
FIGURE 9 14 Light undergoes successive total
internal reflections as it moves through an
optical fibre Rationalised 2023-24
Ray Optics and
Optical Instruments
233
EXAMPLE 9 |
9 | 370-373 | We have, for small angles,
tan ÐNOM = MN
OM
FIGURE 9 14 Light undergoes successive total
internal reflections as it moves through an
optical fibre Rationalised 2023-24
Ray Optics and
Optical Instruments
233
EXAMPLE 9 5
tan ÐNCM = MN
MC
tan ÐNIM = MN
MI
Now, for DNOC, i is the exterior angle |
9 | 371-374 | 14 Light undergoes successive total
internal reflections as it moves through an
optical fibre Rationalised 2023-24
Ray Optics and
Optical Instruments
233
EXAMPLE 9 5
tan ÐNCM = MN
MC
tan ÐNIM = MN
MI
Now, for DNOC, i is the exterior angle Therefore, i
= ÐNOM + ÐNCM
i = MN
MN
OM
+MC
(9 |
9 | 372-375 | Rationalised 2023-24
Ray Optics and
Optical Instruments
233
EXAMPLE 9 5
tan ÐNCM = MN
MC
tan ÐNIM = MN
MI
Now, for DNOC, i is the exterior angle Therefore, i
= ÐNOM + ÐNCM
i = MN
MN
OM
+MC
(9 13)
Similarly,
r = ÐNCM – ÐNIM
i |
9 | 373-376 | 5
tan ÐNCM = MN
MC
tan ÐNIM = MN
MI
Now, for DNOC, i is the exterior angle Therefore, i
= ÐNOM + ÐNCM
i = MN
MN
OM
+MC
(9 13)
Similarly,
r = ÐNCM – ÐNIM
i e |
9 | 374-377 | Therefore, i
= ÐNOM + ÐNCM
i = MN
MN
OM
+MC
(9 13)
Similarly,
r = ÐNCM – ÐNIM
i e , r = MN
MN
MC
MI
−
(9 |
9 | 375-378 | 13)
Similarly,
r = ÐNCM – ÐNIM
i e , r = MN
MN
MC
MI
−
(9 14)
Now, by Snell’s law
n1 sin i = n 2 sin r
or for small angles
n1i = n 2r
Substituting i and r from Eqs |
9 | 376-379 | e , r = MN
MN
MC
MI
−
(9 14)
Now, by Snell’s law
n1 sin i = n 2 sin r
or for small angles
n1i = n 2r
Substituting i and r from Eqs (9 |
9 | 377-380 | , r = MN
MN
MC
MI
−
(9 14)
Now, by Snell’s law
n1 sin i = n 2 sin r
or for small angles
n1i = n 2r
Substituting i and r from Eqs (9 13) and (9 |
9 | 378-381 | 14)
Now, by Snell’s law
n1 sin i = n 2 sin r
or for small angles
n1i = n 2r
Substituting i and r from Eqs (9 13) and (9 14), we get
1
2
2
1
OM
MI
MC
n
n
n
n
−
+
=
(9 |
9 | 379-382 | (9 13) and (9 14), we get
1
2
2
1
OM
MI
MC
n
n
n
n
−
+
=
(9 15)
Here, OM, MI and MC represent magnitudes of distances |
9 | 380-383 | 13) and (9 14), we get
1
2
2
1
OM
MI
MC
n
n
n
n
−
+
=
(9 15)
Here, OM, MI and MC represent magnitudes of distances Applying the
Cartesian sign convention,
OM = –u, MI = +v, MC = +R
Substituting these in Eq |
9 | 381-384 | 14), we get
1
2
2
1
OM
MI
MC
n
n
n
n
−
+
=
(9 15)
Here, OM, MI and MC represent magnitudes of distances Applying the
Cartesian sign convention,
OM = –u, MI = +v, MC = +R
Substituting these in Eq (9 |
9 | 382-385 | 15)
Here, OM, MI and MC represent magnitudes of distances Applying the
Cartesian sign convention,
OM = –u, MI = +v, MC = +R
Substituting these in Eq (9 15), we get
2
1
2
1
n
n
n
n
v
u
−R
−
=
(9 |
9 | 383-386 | Applying the
Cartesian sign convention,
OM = –u, MI = +v, MC = +R
Substituting these in Eq (9 15), we get
2
1
2
1
n
n
n
n
v
u
−R
−
=
(9 16)
Equation (9 |
9 | 384-387 | (9 15), we get
2
1
2
1
n
n
n
n
v
u
−R
−
=
(9 16)
Equation (9 16) gives us a relation between object and image distance
in terms of refractive index of the medium and the radius of
curvature of the curved spherical surface |
9 | 385-388 | 15), we get
2
1
2
1
n
n
n
n
v
u
−R
−
=
(9 16)
Equation (9 16) gives us a relation between object and image distance
in terms of refractive index of the medium and the radius of
curvature of the curved spherical surface It holds for any curved
spherical surface |
9 | 386-389 | 16)
Equation (9 16) gives us a relation between object and image distance
in terms of refractive index of the medium and the radius of
curvature of the curved spherical surface It holds for any curved
spherical surface Example 9 |
9 | 387-390 | 16) gives us a relation between object and image distance
in terms of refractive index of the medium and the radius of
curvature of the curved spherical surface It holds for any curved
spherical surface Example 9 5 Light from a point source in air falls on a spherical
glass surface (n = 1 |
9 | 388-391 | It holds for any curved
spherical surface Example 9 5 Light from a point source in air falls on a spherical
glass surface (n = 1 5 and radius of curvature = 20 cm) |
9 | 389-392 | Example 9 5 Light from a point source in air falls on a spherical
glass surface (n = 1 5 and radius of curvature = 20 cm) The distance
of the light source from the glass surface is 100 cm |
9 | 390-393 | 5 Light from a point source in air falls on a spherical
glass surface (n = 1 5 and radius of curvature = 20 cm) The distance
of the light source from the glass surface is 100 cm At what position
the image is formed |
9 | 391-394 | 5 and radius of curvature = 20 cm) The distance
of the light source from the glass surface is 100 cm At what position
the image is formed Solution
We use the relation given by Eq |
9 | 392-395 | The distance
of the light source from the glass surface is 100 cm At what position
the image is formed Solution
We use the relation given by Eq (9 |
9 | 393-396 | At what position
the image is formed Solution
We use the relation given by Eq (9 16) |
9 | 394-397 | Solution
We use the relation given by Eq (9 16) Here
u = – 100 cm, v = |
9 | 395-398 | (9 16) Here
u = – 100 cm, v = , R = + 20 cm, n1 = 1, and n2 = 1 |
9 | 396-399 | 16) Here
u = – 100 cm, v = , R = + 20 cm, n1 = 1, and n2 = 1 5 |
9 | 397-400 | Here
u = – 100 cm, v = , R = + 20 cm, n1 = 1, and n2 = 1 5 We then have
1 |
9 | 398-401 | , R = + 20 cm, n1 = 1, and n2 = 1 5 We then have
1 5
1
0 |
9 | 399-402 | 5 We then have
1 5
1
0 5
100
20
v +
=
or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface,
in the direction of incident light |
9 | 400-403 | We then have
1 5
1
0 5
100
20
v +
=
or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface,
in the direction of incident light FIGURE 9 |
9 | 401-404 | 5
1
0 5
100
20
v +
=
or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface,
in the direction of incident light FIGURE 9 15 Refraction at a spherical
surface separating two media |
9 | 402-405 | 5
100
20
v +
=
or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface,
in the direction of incident light FIGURE 9 15 Refraction at a spherical
surface separating two media Rationalised 2023-24
Physics
234
9 |
9 | 403-406 | FIGURE 9 15 Refraction at a spherical
surface separating two media Rationalised 2023-24
Physics
234
9 5 |
9 | 404-407 | 15 Refraction at a spherical
surface separating two media Rationalised 2023-24
Physics
234
9 5 2 Refraction by a lens
Figure 9 |
9 | 405-408 | Rationalised 2023-24
Physics
234
9 5 2 Refraction by a lens
Figure 9 16(a) shows the geometry of image formation by a double convex
lens |
9 | 406-409 | 5 2 Refraction by a lens
Figure 9 16(a) shows the geometry of image formation by a double convex
lens The image formation can be seen in terms of two steps:
(i) The first refracting surface forms the image I1 of the object O
[Fig |
9 | 407-410 | 2 Refraction by a lens
Figure 9 16(a) shows the geometry of image formation by a double convex
lens The image formation can be seen in terms of two steps:
(i) The first refracting surface forms the image I1 of the object O
[Fig 9 |
9 | 408-411 | 16(a) shows the geometry of image formation by a double convex
lens The image formation can be seen in terms of two steps:
(i) The first refracting surface forms the image I1 of the object O
[Fig 9 16(b)] |
9 | 409-412 | The image formation can be seen in terms of two steps:
(i) The first refracting surface forms the image I1 of the object O
[Fig 9 16(b)] The image I1 acts as a virtual object for the second surface
that forms the image at I [Fig |
9 | 410-413 | 9 16(b)] The image I1 acts as a virtual object for the second surface
that forms the image at I [Fig 9 |
9 | 411-414 | 16(b)] The image I1 acts as a virtual object for the second surface
that forms the image at I [Fig 9 16(c)] |
9 | 412-415 | The image I1 acts as a virtual object for the second surface
that forms the image at I [Fig 9 16(c)] Applying Eq |
9 | 413-416 | 9 16(c)] Applying Eq (9 |
9 | 414-417 | 16(c)] Applying Eq (9 15) to the first
interface ABC, we get
1
2
2
1
1
1
OB
BI
BC
n
n
n
−n
+
=
(9 |
9 | 415-418 | Applying Eq (9 15) to the first
interface ABC, we get
1
2
2
1
1
1
OB
BI
BC
n
n
n
−n
+
=
(9 17)
A similar procedure applied to the second
interface* ADC gives,
2
1
2
1
1
2
DI
DI
DC
n
n
n
−n
−
+
=
(9 |
9 | 416-419 | (9 15) to the first
interface ABC, we get
1
2
2
1
1
1
OB
BI
BC
n
n
n
−n
+
=
(9 17)
A similar procedure applied to the second
interface* ADC gives,
2
1
2
1
1
2
DI
DI
DC
n
n
n
−n
−
+
=
(9 18)
For a thin lens, BI1 = DI1 |
9 | 417-420 | 15) to the first
interface ABC, we get
1
2
2
1
1
1
OB
BI
BC
n
n
n
−n
+
=
(9 17)
A similar procedure applied to the second
interface* ADC gives,
2
1
2
1
1
2
DI
DI
DC
n
n
n
−n
−
+
=
(9 18)
For a thin lens, BI1 = DI1 Adding
Eqs |
9 | 418-421 | 17)
A similar procedure applied to the second
interface* ADC gives,
2
1
2
1
1
2
DI
DI
DC
n
n
n
−n
−
+
=
(9 18)
For a thin lens, BI1 = DI1 Adding
Eqs (9 |
9 | 419-422 | 18)
For a thin lens, BI1 = DI1 Adding
Eqs (9 17) and (9 |
9 | 420-423 | Adding
Eqs (9 17) and (9 18), we get
n
n
n
n
1
1
2
1
1
1
OB
DI
BC
DC
1
2
+
=
−
+
(
)
(9 |
9 | 421-424 | (9 17) and (9 18), we get
n
n
n
n
1
1
2
1
1
1
OB
DI
BC
DC
1
2
+
=
−
+
(
)
(9 19)
Suppose the object is at infinity, i |
9 | 422-425 | 17) and (9 18), we get
n
n
n
n
1
1
2
1
1
1
OB
DI
BC
DC
1
2
+
=
−
+
(
)
(9 19)
Suppose the object is at infinity, i e |
9 | 423-426 | 18), we get
n
n
n
n
1
1
2
1
1
1
OB
DI
BC
DC
1
2
+
=
−
+
(
)
(9 19)
Suppose the object is at infinity, i e ,
OB ® ¥ and DI = f, Eq |
9 | 424-427 | 19)
Suppose the object is at infinity, i e ,
OB ® ¥ and DI = f, Eq (9 |
9 | 425-428 | e ,
OB ® ¥ and DI = f, Eq (9 19) gives
fn
n
n
1
2
1
1
1
=
−
+
(
)
BC
DC
1
2
(9 |
9 | 426-429 | ,
OB ® ¥ and DI = f, Eq (9 19) gives
fn
n
n
1
2
1
1
1
=
−
+
(
)
BC
DC
1
2
(9 20)
The point where image of an object
placed at infinity is formed is called the
focus F, of the lens and the distance f gives
its focal length |
9 | 427-430 | (9 19) gives
fn
n
n
1
2
1
1
1
=
−
+
(
)
BC
DC
1
2
(9 20)
The point where image of an object
placed at infinity is formed is called the
focus F, of the lens and the distance f gives
its focal length A lens has two foci, F and
F¢, on either side of it (Fig |
9 | 428-431 | 19) gives
fn
n
n
1
2
1
1
1
=
−
+
(
)
BC
DC
1
2
(9 20)
The point where image of an object
placed at infinity is formed is called the
focus F, of the lens and the distance f gives
its focal length A lens has two foci, F and
F¢, on either side of it (Fig 9 |
9 | 429-432 | 20)
The point where image of an object
placed at infinity is formed is called the
focus F, of the lens and the distance f gives
its focal length A lens has two foci, F and
F¢, on either side of it (Fig 9 16) |
9 | 430-433 | A lens has two foci, F and
F¢, on either side of it (Fig 9 16) By the
sign convention,
BC1 = + R1,
DC2 = –R2
So Eq |
9 | 431-434 | 9 16) By the
sign convention,
BC1 = + R1,
DC2 = –R2
So Eq (9 |
9 | 432-435 | 16) By the
sign convention,
BC1 = + R1,
DC2 = –R2
So Eq (9 20) can be written as
(9 |
9 | 433-436 | By the
sign convention,
BC1 = + R1,
DC2 = –R2
So Eq (9 20) can be written as
(9 21)
Equation (9 |
9 | 434-437 | (9 20) can be written as
(9 21)
Equation (9 21) is known as the lens
maker’s formula |
9 | 435-438 | 20) can be written as
(9 21)
Equation (9 21) is known as the lens
maker’s formula It is useful to design
lenses of desired focal length using surfaces
of suitable radii of curvature |
9 | 436-439 | 21)
Equation (9 21) is known as the lens
maker’s formula It is useful to design
lenses of desired focal length using surfaces
of suitable radii of curvature Note that the
formula is true for a concave lens also |
9 | 437-440 | 21) is known as the lens
maker’s formula It is useful to design
lenses of desired focal length using surfaces
of suitable radii of curvature Note that the
formula is true for a concave lens also In
that case R1is negative, R 2 positive and
therefore, f is negative |
9 | 438-441 | It is useful to design
lenses of desired focal length using surfaces
of suitable radii of curvature 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 |
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