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9 | 3839-3842 | 0 cm It can be seen with an explicit ray
diagram that the answer is independent of the location of the slab
(for small angles of incidence) 9 17
(a)
sin i¢c = 1 |
9 | 3840-3843 | It can be seen with an explicit ray
diagram that the answer is independent of the location of the slab
(for small angles of incidence) 9 17
(a)
sin i¢c = 1 44/1 |
9 | 3841-3844 | 9 17
(a)
sin i¢c = 1 44/1 68 which gives i¢c = 59° |
9 | 3842-3845 | 17
(a)
sin i¢c = 1 44/1 68 which gives i¢c = 59° Total internal reflection
takes place when i > 59° or when r < rmax = 31° |
9 | 3843-3846 | 44/1 68 which gives i¢c = 59° Total internal reflection
takes place when i > 59° or when r < rmax = 31° Now,
(sin
/sin
) |
9 | 3844-3847 | 68 which gives i¢c = 59° Total internal reflection
takes place when i > 59° or when r < rmax = 31° Now,
(sin
/sin
) max
max
i
r
= 1 68 , which gives imax ~ 60° |
9 | 3845-3848 | Total internal reflection
takes place when i > 59° or when r < rmax = 31° Now,
(sin
/sin
) max
max
i
r
= 1 68 , which gives imax ~ 60° Thus, all
incident rays of angles in the range 0 < i < 60° will suffer total
internal reflections in the pipe |
9 | 3846-3849 | Now,
(sin
/sin
) max
max
i
r
= 1 68 , which gives imax ~ 60° Thus, all
incident rays of angles in the range 0 < i < 60° will suffer total
internal reflections in the pipe (If the length of the pipe is
finite, which it is in practice, there will be a lower limit on i
determined by the ratio of the diameter to the length of the
pipe |
9 | 3847-3850 | max
max
i
r
= 1 68 , which gives imax ~ 60° Thus, all
incident rays of angles in the range 0 < i < 60° will suffer total
internal reflections in the pipe (If the length of the pipe is
finite, which it is in practice, there will be a lower limit on i
determined by the ratio of the diameter to the length of the
pipe )
(b)
If there is no outer coating, i¢c = sin–1(1/1 |
9 | 3848-3851 | Thus, all
incident rays of angles in the range 0 < i < 60° will suffer total
internal reflections in the pipe (If the length of the pipe is
finite, which it is in practice, there will be a lower limit on i
determined by the ratio of the diameter to the length of the
pipe )
(b)
If there is no outer coating, i¢c = sin–1(1/1 68) = 36 |
9 | 3849-3852 | (If the length of the pipe is
finite, which it is in practice, there will be a lower limit on i
determined by the ratio of the diameter to the length of the
pipe )
(b)
If there is no outer coating, i¢c = sin–1(1/1 68) = 36 5° |
9 | 3850-3853 | )
(b)
If there is no outer coating, i¢c = sin–1(1/1 68) = 36 5° Now,
i = 90° will have r = 36 |
9 | 3851-3854 | 68) = 36 5° Now,
i = 90° will have r = 36 5° and i¢ = 53 |
9 | 3852-3855 | 5° Now,
i = 90° will have r = 36 5° and i¢ = 53 5° which is greater than
i¢c |
9 | 3853-3856 | Now,
i = 90° will have r = 36 5° and i¢ = 53 5° which is greater than
i¢c Thus, all incident rays (in the range 53 |
9 | 3854-3857 | 5° and i¢ = 53 5° which is greater than
i¢c Thus, all incident rays (in the range 53 5° < i < 90°) will
suffer total internal reflections |
9 | 3855-3858 | 5° which is greater than
i¢c Thus, all incident rays (in the range 53 5° < i < 90°) will
suffer total internal reflections 9 |
9 | 3856-3859 | Thus, all incident rays (in the range 53 5° < i < 90°) will
suffer total internal reflections 9 18
For fixed distance s between object and screen, the lens equation
does not give a real solution for u or v if f is greater than s/4 |
9 | 3857-3860 | 5° < i < 90°) will
suffer total internal reflections 9 18
For fixed distance s between object and screen, the lens equation
does not give a real solution for u or v if f is greater than s/4 Therefore, fmax = 0 |
9 | 3858-3861 | 9 18
For fixed distance s between object and screen, the lens equation
does not give a real solution for u or v if f is greater than s/4 Therefore, fmax = 0 75 m |
9 | 3859-3862 | 18
For fixed distance s between object and screen, the lens equation
does not give a real solution for u or v if f is greater than s/4 Therefore, fmax = 0 75 m 9 |
9 | 3860-3863 | Therefore, fmax = 0 75 m 9 19
21 |
9 | 3861-3864 | 75 m 9 19
21 4 cm
9 |
9 | 3862-3865 | 9 19
21 4 cm
9 20
(a)
(i) Let a parallel beam be the incident from the left on the convex
lens first |
9 | 3863-3866 | 19
21 4 cm
9 20
(a)
(i) Let a parallel beam be the incident from the left on the convex
lens first f1 = 30 cm and u1 = – , give v1 = + 30 cm |
9 | 3864-3867 | 4 cm
9 20
(a)
(i) Let a parallel beam be the incident from the left on the convex
lens first f1 = 30 cm and u1 = – , give v1 = + 30 cm This image becomes
a virtual object for the second lens |
9 | 3865-3868 | 20
(a)
(i) Let a parallel beam be the incident from the left on the convex
lens first f1 = 30 cm and u1 = – , give v1 = + 30 cm This image becomes
a virtual object for the second lens f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives,
v2 = – 220 cm |
9 | 3866-3869 | f1 = 30 cm and u1 = – , give v1 = + 30 cm This image becomes
a virtual object for the second lens f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives,
v2 = – 220 cm The parallel incident beam appears to diverge
from a point 216 cm from the centre of the two-lens system |
9 | 3867-3870 | This image becomes
a virtual object for the second lens f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives,
v2 = – 220 cm The parallel incident beam appears to diverge
from a point 216 cm from the centre of the two-lens system (ii) Let the parallel beam be incident from the left on the concave
lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm |
9 | 3868-3871 | f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives,
v2 = – 220 cm The parallel incident beam appears to diverge
from a point 216 cm from the centre of the two-lens system (ii) Let the parallel beam be incident from the left on the concave
lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm This image
becomes a real object for the second lens: f2 = + 30 cm, u2 =
– (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm |
9 | 3869-3872 | The parallel incident beam appears to diverge
from a point 216 cm from the centre of the two-lens system (ii) Let the parallel beam be incident from the left on the concave
lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm This image
becomes a real object for the second lens: f2 = + 30 cm, u2 =
– (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm The parallel
incident beam appears to diverge from a point 416 cm on the
left of the centre of the two-lens system |
9 | 3870-3873 | (ii) Let the parallel beam be incident from the left on the concave
lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm This image
becomes a real object for the second lens: f2 = + 30 cm, u2 =
– (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm The parallel
incident beam appears to diverge from a point 416 cm on the
left of the centre of the two-lens system Clearly, the answer depends on which side of the lens system
the parallel beam is incident |
9 | 3871-3874 | This image
becomes a real object for the second lens: f2 = + 30 cm, u2 =
– (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm The parallel
incident beam appears to diverge from a point 416 cm on the
left of the centre of the two-lens system Clearly, the answer depends on which side of the lens system
the parallel beam is incident Further we do not have a simple lens
equation true for all u (and v) in terms of a definite constant of the
system (the constant being determined by f1 and f2, and the separation
between the lenses) |
9 | 3872-3875 | The parallel
incident beam appears to diverge from a point 416 cm on the
left of the centre of the two-lens system Clearly, the answer depends on which side of the lens system
the parallel beam is incident Further we do not have a simple lens
equation true for all u (and v) in terms of a definite constant of the
system (the constant being determined by f1 and f2, and the separation
between the lenses) The notion of effective focal length, therefore,
does not seem to be meaningful for this system |
9 | 3873-3876 | Clearly, the answer depends on which side of the lens system
the parallel beam is incident Further we do not have a simple lens
equation true for all u (and v) in terms of a definite constant of the
system (the constant being determined by f1 and f2, and the separation
between the lenses) The notion of effective focal length, therefore,
does not seem to be meaningful for this system (b)
u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm |
9 | 3874-3877 | Further we do not have a simple lens
equation true for all u (and v) in terms of a definite constant of the
system (the constant being determined by f1 and f2, and the separation
between the lenses) The notion of effective focal length, therefore,
does not seem to be meaningful for this system (b)
u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm Magnitude of magnification due to the first (convex) lens is 3 |
9 | 3875-3878 | The notion of effective focal length, therefore,
does not seem to be meaningful for this system (b)
u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm Magnitude of magnification due to the first (convex) lens is 3 u 2 = + (120 – 8) cm = +112 cm (object virtual);
f2 = – 20 cm which gives v2
112
9220
= −
×
cm
Magnitude of magnification due to the second (concave)
Rationalised 2023-24
348
Physics
lens = 20/92 |
9 | 3876-3879 | (b)
u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm Magnitude of magnification due to the first (convex) lens is 3 u 2 = + (120 – 8) cm = +112 cm (object virtual);
f2 = – 20 cm which gives v2
112
9220
= −
×
cm
Magnitude of magnification due to the second (concave)
Rationalised 2023-24
348
Physics
lens = 20/92 Net magnitude of magnification = 0 |
9 | 3877-3880 | Magnitude of magnification due to the first (convex) lens is 3 u 2 = + (120 – 8) cm = +112 cm (object virtual);
f2 = – 20 cm which gives v2
112
9220
= −
×
cm
Magnitude of magnification due to the second (concave)
Rationalised 2023-24
348
Physics
lens = 20/92 Net magnitude of magnification = 0 652
Size of the image = 0 |
9 | 3878-3881 | u 2 = + (120 – 8) cm = +112 cm (object virtual);
f2 = – 20 cm which gives v2
112
9220
= −
×
cm
Magnitude of magnification due to the second (concave)
Rationalised 2023-24
348
Physics
lens = 20/92 Net magnitude of magnification = 0 652
Size of the image = 0 98 cm
9 |
9 | 3879-3882 | Net magnitude of magnification = 0 652
Size of the image = 0 98 cm
9 21
If the refracted ray in the prism is incident on the second face at the
critical angle ic, the angle of refraction r at the first face is (60°–ic) |
9 | 3880-3883 | 652
Size of the image = 0 98 cm
9 21
If the refracted ray in the prism is incident on the second face at the
critical angle ic, the angle of refraction r at the first face is (60°–ic) Now, ic = sin–1 (1/1 |
9 | 3881-3884 | 98 cm
9 21
If the refracted ray in the prism is incident on the second face at the
critical angle ic, the angle of refraction r at the first face is (60°–ic) Now, ic = sin–1 (1/1 524) ~ 41°
Therefore, r = 19°
sin i = 0 |
9 | 3882-3885 | 21
If the refracted ray in the prism is incident on the second face at the
critical angle ic, the angle of refraction r at the first face is (60°–ic) Now, ic = sin–1 (1/1 524) ~ 41°
Therefore, r = 19°
sin i = 0 4962; i ~ 30°
9 |
9 | 3883-3886 | Now, ic = sin–1 (1/1 524) ~ 41°
Therefore, r = 19°
sin i = 0 4962; i ~ 30°
9 22
(a)
1
91
1
10
v
+
=
i |
9 | 3884-3887 | 524) ~ 41°
Therefore, r = 19°
sin i = 0 4962; i ~ 30°
9 22
(a)
1
91
1
10
v
+
=
i e |
9 | 3885-3888 | 4962; i ~ 30°
9 22
(a)
1
91
1
10
v
+
=
i e ,
v = – 90 cm,
Magnitude of magnification = 90/9 = 10 |
9 | 3886-3889 | 22
(a)
1
91
1
10
v
+
=
i e ,
v = – 90 cm,
Magnitude of magnification = 90/9 = 10 Each square in the virtual image has an area 10 × 10 × 1 mm2
= 100 mm2 = 1 cm2
(b)
Magnifying power = 25/9 = 2 |
9 | 3887-3890 | e ,
v = – 90 cm,
Magnitude of magnification = 90/9 = 10 Each square in the virtual image has an area 10 × 10 × 1 mm2
= 100 mm2 = 1 cm2
(b)
Magnifying power = 25/9 = 2 8
(c)
No, magnification of an image by a lens and angular magnification
(or magnifying power) of an optical instrument are two separate
things |
9 | 3888-3891 | ,
v = – 90 cm,
Magnitude of magnification = 90/9 = 10 Each square in the virtual image has an area 10 × 10 × 1 mm2
= 100 mm2 = 1 cm2
(b)
Magnifying power = 25/9 = 2 8
(c)
No, magnification of an image by a lens and angular magnification
(or magnifying power) of an optical instrument are two separate
things The latter is the ratio of the angular size of the object
(which is equal to the angular size of the image even if the image
is magnified) to the angular size of the object if placed at the near
point (25 cm) |
9 | 3889-3892 | Each square in the virtual image has an area 10 × 10 × 1 mm2
= 100 mm2 = 1 cm2
(b)
Magnifying power = 25/9 = 2 8
(c)
No, magnification of an image by a lens and angular magnification
(or magnifying power) of an optical instrument are two separate
things The latter is the ratio of the angular size of the object
(which is equal to the angular size of the image even if the image
is magnified) to the angular size of the object if placed at the near
point (25 cm) Thus, magnification magnitude is |(v/u)| and
magnifying power is (25/ |u|) |
9 | 3890-3893 | 8
(c)
No, magnification of an image by a lens and angular magnification
(or magnifying power) of an optical instrument are two separate
things The latter is the ratio of the angular size of the object
(which is equal to the angular size of the image even if the image
is magnified) to the angular size of the object if placed at the near
point (25 cm) Thus, magnification magnitude is |(v/u)| and
magnifying power is (25/ |u|) Only when the image is located at
the near point |v| = 25 cm, are the two quantities equal |
9 | 3891-3894 | The latter is the ratio of the angular size of the object
(which is equal to the angular size of the image even if the image
is magnified) to the angular size of the object if placed at the near
point (25 cm) Thus, magnification magnitude is |(v/u)| and
magnifying power is (25/ |u|) Only when the image is located at
the near point |v| = 25 cm, are the two quantities equal 9 |
9 | 3892-3895 | Thus, magnification magnitude is |(v/u)| and
magnifying power is (25/ |u|) Only when the image is located at
the near point |v| = 25 cm, are the two quantities equal 9 23
(a)
Maximum magnifying power is obtained when the image is at
the near point (25 cm)
u = – 7 |
9 | 3893-3896 | Only when the image is located at
the near point |v| = 25 cm, are the two quantities equal 9 23
(a)
Maximum magnifying power is obtained when the image is at
the near point (25 cm)
u = – 7 14 cm |
9 | 3894-3897 | 9 23
(a)
Maximum magnifying power is obtained when the image is at
the near point (25 cm)
u = – 7 14 cm (b)
Magnitude of magnification = (25/ |u|) = 3 |
9 | 3895-3898 | 23
(a)
Maximum magnifying power is obtained when the image is at
the near point (25 cm)
u = – 7 14 cm (b)
Magnitude of magnification = (25/ |u|) = 3 5 |
9 | 3896-3899 | 14 cm (b)
Magnitude of magnification = (25/ |u|) = 3 5 (c)
Magnifying power = 3 |
9 | 3897-3900 | (b)
Magnitude of magnification = (25/ |u|) = 3 5 (c)
Magnifying power = 3 5
Yes, the magnifying power (when the image is produced at 25 cm)
is equal to the magnitude of magnification |
9 | 3898-3901 | 5 (c)
Magnifying power = 3 5
Yes, the magnifying power (when the image is produced at 25 cm)
is equal to the magnitude of magnification 9 |
9 | 3899-3902 | (c)
Magnifying power = 3 5
Yes, the magnifying power (when the image is produced at 25 cm)
is equal to the magnitude of magnification 9 24
Magnification =
( |
9 | 3900-3903 | 5
Yes, the magnifying power (when the image is produced at 25 cm)
is equal to the magnitude of magnification 9 24
Magnification =
( 6 25/ )
1 = 2 |
9 | 3901-3904 | 9 24
Magnification =
( 6 25/ )
1 = 2 5
v = +2 |
9 | 3902-3905 | 24
Magnification =
( 6 25/ )
1 = 2 5
v = +2 5u
1
2 5
1
1
10 |
9 | 3903-3906 | 6 25/ )
1 = 2 5
v = +2 5u
1
2 5
1
1
10 u
u
i |
9 | 3904-3907 | 5
v = +2 5u
1
2 5
1
1
10 u
u
i e |
9 | 3905-3908 | 5u
1
2 5
1
1
10 u
u
i e ,u = – 6 cm
|v| = 15 cm
The virtual image is closer than the normal near point (25 cm) and
cannot be seen by the eye distinctly |
9 | 3906-3909 | u
u
i e ,u = – 6 cm
|v| = 15 cm
The virtual image is closer than the normal near point (25 cm) and
cannot be seen by the eye distinctly 9 |
9 | 3907-3910 | e ,u = – 6 cm
|v| = 15 cm
The virtual image is closer than the normal near point (25 cm) and
cannot be seen by the eye distinctly 9 25
(a)
Even though the absolute image size is bigger than the object
size, the angular size of the image is equal to the angular size of
the object |
9 | 3908-3911 | ,u = – 6 cm
|v| = 15 cm
The virtual image is closer than the normal near point (25 cm) and
cannot be seen by the eye distinctly 9 25
(a)
Even though the absolute image size is bigger than the object
size, the angular size of the image is equal to the angular size of
the object The magnifier helps in the following way: without it
object would be placed no closer than 25 cm; with it the object
can be placed much closer |
9 | 3909-3912 | 9 25
(a)
Even though the absolute image size is bigger than the object
size, the angular size of the image is equal to the angular size of
the object The magnifier helps in the following way: without it
object would be placed no closer than 25 cm; with it the object
can be placed much closer The closer object has larger angular
size than the same object at 25 cm |
9 | 3910-3913 | 25
(a)
Even though the absolute image size is bigger than the object
size, the angular size of the image is equal to the angular size of
the object The magnifier helps in the following way: without it
object would be placed no closer than 25 cm; with it the object
can be placed much closer The closer object has larger angular
size than the same object at 25 cm It is in this sense that angular
magnification is achieved |
9 | 3911-3914 | The magnifier helps in the following way: without it
object would be placed no closer than 25 cm; with it the object
can be placed much closer The closer object has larger angular
size than the same object at 25 cm It is in this sense that angular
magnification is achieved (b)
Yes, it decreases a little because the angle subtended at the eye
is then slightly less than the angle subtended at the lens |
9 | 3912-3915 | The closer object has larger angular
size than the same object at 25 cm It is in this sense that angular
magnification is achieved (b)
Yes, it decreases a little because the angle subtended at the eye
is then slightly less than the angle subtended at the lens The
Rationalised 2023-24
349
Answers
effect is negligible if the image is at a very large distance away |
9 | 3913-3916 | It is in this sense that angular
magnification is achieved (b)
Yes, it decreases a little because the angle subtended at the eye
is then slightly less than the angle subtended at the lens The
Rationalised 2023-24
349
Answers
effect is negligible if the image is at a very large distance away [Note: When the eye is separated from the lens, the angles
subtended at the eye by the first object and its image are not
equal |
9 | 3914-3917 | (b)
Yes, it decreases a little because the angle subtended at the eye
is then slightly less than the angle subtended at the lens The
Rationalised 2023-24
349
Answers
effect is negligible if the image is at a very large distance away [Note: When the eye is separated from the lens, the angles
subtended at the eye by the first object and its image are not
equal ]
(c)
First, grinding lens of very small focal length is not easy |
9 | 3915-3918 | The
Rationalised 2023-24
349
Answers
effect is negligible if the image is at a very large distance away [Note: When the eye is separated from the lens, the angles
subtended at the eye by the first object and its image are not
equal ]
(c)
First, grinding lens of very small focal length is not easy More
important, if you decrease focal length, aberrations (both spherical
and chromatic) become more pronounced |
9 | 3916-3919 | [Note: When the eye is separated from the lens, the angles
subtended at the eye by the first object and its image are not
equal ]
(c)
First, grinding lens of very small focal length is not easy More
important, if you decrease focal length, aberrations (both spherical
and chromatic) become more pronounced So, in practice, you
cannot get a magnifying power of more than 3 or so with a simple
convex lens |
9 | 3917-3920 | ]
(c)
First, grinding lens of very small focal length is not easy More
important, if you decrease focal length, aberrations (both spherical
and chromatic) become more pronounced So, in practice, you
cannot get a magnifying power of more than 3 or so with a simple
convex lens However, using an aberration corrected lens system,
one can increase this limit by a factor of 10 or so |
9 | 3918-3921 | More
important, if you decrease focal length, aberrations (both spherical
and chromatic) become more pronounced So, in practice, you
cannot get a magnifying power of more than 3 or so with a simple
convex lens However, using an aberration corrected lens system,
one can increase this limit by a factor of 10 or so (d)
Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which
increases if fe is smaller |
9 | 3919-3922 | So, in practice, you
cannot get a magnifying power of more than 3 or so with a simple
convex lens However, using an aberration corrected lens system,
one can increase this limit by a factor of 10 or so (d)
Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which
increases if fe is smaller Further, magnification of the objective
is given by
O
O
O
O
1
|
|
(|
|/
)
1
v
u
u
f
=
−
which is large when
|O
u|
is slightly greater than fO |
9 | 3920-3923 | However, using an aberration corrected lens system,
one can increase this limit by a factor of 10 or so (d)
Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which
increases if fe is smaller Further, magnification of the objective
is given by
O
O
O
O
1
|
|
(|
|/
)
1
v
u
u
f
=
−
which is large when
|O
u|
is slightly greater than fO The micro-
scope is used for viewing very close object |
9 | 3921-3924 | (d)
Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which
increases if fe is smaller Further, magnification of the objective
is given by
O
O
O
O
1
|
|
(|
|/
)
1
v
u
u
f
=
−
which is large when
|O
u|
is slightly greater than fO The micro-
scope is used for viewing very close object So
|O
u|
is small, and
so is fO |
9 | 3922-3925 | Further, magnification of the objective
is given by
O
O
O
O
1
|
|
(|
|/
)
1
v
u
u
f
=
−
which is large when
|O
u|
is slightly greater than fO The micro-
scope is used for viewing very close object So
|O
u|
is small, and
so is fO (e)
The image of the objective in the eye-piece is known as ‘eye-ring’ |
9 | 3923-3926 | The micro-
scope is used for viewing very close object So
|O
u|
is small, and
so is fO (e)
The image of the objective in the eye-piece is known as ‘eye-ring’ All the rays from the object refracted by objective go through the
eye-ring |
9 | 3924-3927 | So
|O
u|
is small, and
so is fO (e)
The image of the objective in the eye-piece is known as ‘eye-ring’ All the rays from the object refracted by objective go through the
eye-ring Therefore, it is an ideal position for our eyes for viewing |
9 | 3925-3928 | (e)
The image of the objective in the eye-piece is known as ‘eye-ring’ All the rays from the object refracted by objective go through the
eye-ring Therefore, it is an ideal position for our eyes for viewing If we place our eyes too close to the eye-piece, we shall not collect
much of the light and also reduce our field of view |
9 | 3926-3929 | All the rays from the object refracted by objective go through the
eye-ring Therefore, it is an ideal position for our eyes for viewing If we place our eyes too close to the eye-piece, we shall not collect
much of the light and also reduce our field of view If we position
our eyes on the eye-ring and the area of the pupil of our eye is
greater or equal to the area of the eye-ring, our eyes will collect
all the light refracted by the objective |
9 | 3927-3930 | Therefore, it is an ideal position for our eyes for viewing If we place our eyes too close to the eye-piece, we shall not collect
much of the light and also reduce our field of view If we position
our eyes on the eye-ring and the area of the pupil of our eye is
greater or equal to the area of the eye-ring, our eyes will collect
all the light refracted by the objective The precise location of
the eye-ring naturally depends on the separation between the
objective and the eye-piece |
9 | 3928-3931 | If we place our eyes too close to the eye-piece, we shall not collect
much of the light and also reduce our field of view If we position
our eyes on the eye-ring and the area of the pupil of our eye is
greater or equal to the area of the eye-ring, our eyes will collect
all the light refracted by the objective The precise location of
the eye-ring naturally depends on the separation between the
objective and the eye-piece When you view through a microscope
by placing your eyes on one end,the ideal distance between the
eyes and eye-piece is usually built-in the design of the
instrument |
9 | 3929-3932 | If we position
our eyes on the eye-ring and the area of the pupil of our eye is
greater or equal to the area of the eye-ring, our eyes will collect
all the light refracted by the objective The precise location of
the eye-ring naturally depends on the separation between the
objective and the eye-piece When you view through a microscope
by placing your eyes on one end,the ideal distance between the
eyes and eye-piece is usually built-in the design of the
instrument 9 |
9 | 3930-3933 | The precise location of
the eye-ring naturally depends on the separation between the
objective and the eye-piece When you view through a microscope
by placing your eyes on one end,the ideal distance between the
eyes and eye-piece is usually built-in the design of the
instrument 9 26
Assume microscope in normal use i |
9 | 3931-3934 | When you view through a microscope
by placing your eyes on one end,the ideal distance between the
eyes and eye-piece is usually built-in the design of the
instrument 9 26
Assume microscope in normal use i e |
9 | 3932-3935 | 9 26
Assume microscope in normal use i e , image at 25 cm |
9 | 3933-3936 | 26
Assume microscope in normal use i e , image at 25 cm Angular
magnification of the eye-piece
= 25
5
1
6
Magnification of the objective
= 30
6
5
O
O
1
1
1
5
1 |
9 | 3934-3937 | e , image at 25 cm Angular
magnification of the eye-piece
= 25
5
1
6
Magnification of the objective
= 30
6
5
O
O
1
1
1
5
1 25
u
−u
=
which gives uO= –1 |
9 | 3935-3938 | , image at 25 cm Angular
magnification of the eye-piece
= 25
5
1
6
Magnification of the objective
= 30
6
5
O
O
1
1
1
5
1 25
u
−u
=
which gives uO= –1 5 cm; v0= 7 |
9 | 3936-3939 | Angular
magnification of the eye-piece
= 25
5
1
6
Magnification of the objective
= 30
6
5
O
O
1
1
1
5
1 25
u
−u
=
which gives uO= –1 5 cm; v0= 7 5 cm |
9 | 3937-3940 | 25
u
−u
=
which gives uO= –1 5 cm; v0= 7 5 cm |
|
ue (25/6) cm = 4 |
9 | 3938-3941 | 5 cm; v0= 7 5 cm |
|
ue (25/6) cm = 4 17 cm |
Subsets and Splits