Download F1 The ray diagram for a simple astronomical refracting telescope is

F1
The ray diagram for
a simple astronomical
refracting telescope is
fo
fe
shown in Figure 23.
The focal points of the
objective and the eyepiece
Fe, Fo
are normally made to
θo
θI
h
coincide so that an inverted,
magnified image is viewed
at infinity. This means that
the observer’s eye is fully
relaxed and subject to less
strain. The angular limit of
virtual image
objective lens
resolution for a telescope
at infinity
eyepiece lens
objective of diameter d is
1.22 λ/d where λ is the Figure 23 A simple astronomical telescope. The objective has a long focal length whilst
wavelength
of
t h e that of the eyepiece is much shorter.
radiation — the larger the
objective diameter, the greater the ability to resolve fine detail on the object. You will find a discussion of
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angular resolving power in Subsection 2.3.
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F2
An achromatic doublet is formed by cementing together two lenses with
equal and opposite chromatic aberrations (see Figure 10). One lens is
converging and made from crown glass, with a low average index and a
low dispersive power; the other is diverging and made from flint glass,
with a high average index and a high dispersive power. The shapes of
the lenses are chosen so that minimum spherical aberration is produced
for zero coma. The cement used is Canada balsam, which is a resin with
refractive index approximately equal to the average value of the two
glasses.
low
dispersive
power
high
dispersive
power
converging
lens crown
glass
diverging
lens flint
glass
Canada balsam
Figure 10
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An achromatic doublet.
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F3
The transverse magnification M tran is the ratio of transverse image size h′ to transverse object size h ,
i.e. Mtran = h′/h.
The angular magnification M ang is the ratio of the angle θ0′ subtended at the observer’s eye by the image to the
angle θ subtended there by the object, i.e. Mang = θ0′/θ.
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The magnifying power Mpower is the ratio of the angle θ0′ subtended by the image to the angle θ0D subtended by
the object when at the eye’s near point (or least distance of distinct vision), i.e. Mpower = θ0′/θD.
(a) For a microscope we compare the angular sizes of the image seen through the microscope, with the ‘best’
(i.e. largest) view we can obtain of the object when it is seen directly, which will be when it is at the near point.
The magnifying power is therefore the term used to describe the performance of a microscope.
(b) A film projector projects a greatly enlarged image of the film so as to fill the screen. The transverse
dimensions are therefore important, so transverse magnification is the quantity used.
(c) The object being viewed in a telescope is usually at a large distance which cannot be reduced. Normally, we
adjust the eyepiece for the relaxed eye so that the image is at infinity and transverse dimension is a meaningless
term; angular magnification therefore describes the telescope’s performance.
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Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route
through the module and to proceed directly to Ready to study? in Subsection 1.3.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the
Closing items.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it
here.
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R1
Three rays are considered: (i) the axial ray which travels along the optical axis from the bottom of the object and
passes undeflected through the centre of the lens; (ii) a ray parallel to the optical axis from the top of the object
which passes through the lens and is refracted so as to cross the optical axis at the focal point F (distance f from
the lens); (iii) a second ray from the top of the object which passes undeviated through the centre of the
thin lens. The top of the image is at the point where rays (ii) and (iii) meet.
Using this method an inverted and enlarged real image is formed beyond 2f.
(In fact the image is inverted, 4 cm in height and 24 cm from the lens.)
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R2
Using the Cartesian sign convention: (1/v) − (1/u) = 1/f which, for u = −4 cm and f = +6 cm, becomes
4(1/v) − 1/(−4 cm) = 1/(6 cm)
i.e. 1/v = (1/6 − 1/4) cm−1 = − 1/(12 cm)4
so4v = −12 cm
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1
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So, the image is positioned 12 cm from the lens. Such an image will be virtual and erect.
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R3
The image is constructed using three rays as before but this time the paraxial ray from the top of the object
diverges and so appears to come from the first focal point on the same side of the lens as the object — a dotted
construction line is drawn back to this point where it meets the axial ray. The third ray is drawn from the top of
the object passing undeviated through the centre of this thin lens. The eye sees a virtual image produced at the
intersection of this third ray and the construction line since that is the point from which these rays appear to
come. A careful diagram gives an image 1.4 cm in height, near to the calculated distance from the lens of 5.5 cm.
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R4
In the dispersion of white light by a glass prism the violet end of the spectrum is refracted more than the red.
The reason for the deviation is that the light is slowed down on entering the glass, since glass has a
refractive index which is greater than that of air (refractive index = speed of light in vacuum/speed of light in the
medium concerned). Consequently, a glass of large refractive index causes greater refraction (or deviation) than
one of low refractive index. Since violet light is deviated more than red light, the refractive index of the glass
must be larger for the shorter wavelengths of violet light than for the longer wavelengths of red light.
Common mistakes here are to reverse the order of the spectrum or to show dispersion only taking place at the
second face. Dispersion begins at the first air/glass interface and the different coloured rays are already separated
when they reach the second face.
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R5
In constructing the ray diagram to show image formation by a concave mirror, three rays are used. The first of
these is the axial ray, (i.e. from the bottom of the object along the optical axis of the mirror), which is simply
reflected back on itself, showing that the bottom of the image is on the axis. A second ray, from the top of the
object, passes through the focal point, strikes the mirror and reflects back parallel to the optical axis.
The third ray, from the top of the object, strikes the point where the optical axis meets the mirror, where it is
reflected back at the same angle beneath the axis. The top of the image is situated where the second and third
rays intersect. The remainder of the image will be perpendicular to the optical axis.
In this problem, the initial object distance (20 cm) exceeds 2f (15 cm) and the above construction then shows that
the image is real, diminished, inverted and situated between f and 2f. If the object is moved along the axis
towards the mirror then the inverted image becomes the same size as the object, when it is at 2f, and is located
directly under it. In its final position, (10 cm) the inverted image is enlarged.
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(If the object were to reach f, 7.5 cm from the mirror, then the inverted image would be located at infinity.)
If you feel unsure of any of the terms referred to in Questions R1 to R5, consult the Glossary.
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T1
The diameter of a five pence piece is about 1 cm and the angular size is the actual diameter divided by the
distance, so we have:
Earth−Moon distance
1m
= −2
Moon diameter
10 m
1
The distance to the Moon is then: 3.5 × 106 × 102 m= 3.5 × 108 m.
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T2
The larger the refractive index of glass, the larger the angle through which light of a given wavelength is
deviated or refracted when crossing an air–glass boundary (at non-normal incidence). As violet light is refracted
through a larger angle than red light, the refractive index of the glass must be larger for the shorter wavelengths
of violet light than for the longer wavelengths of red light. Since refractive index is the ratio of the speed of light
in vacuo to the speed in the medium concerned, a large index means a low speed and therefore the violet light
travels more slowly in the glass than the red light.
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T3
(a) For the eye: θ = (1.22 × 5 × 10−7 m)/(5 × 10−3 m)2= 1.22 × 10−4 rad
(b) For the binoculars (taking a typical front lens diameter of 5 cm)
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θ = (1.22 × 5 ×
10−71m)/(5
×
10−21m)
= 1.22 ×
10−51rad
(c) For the optical telescope
θ = (1.22 × 5 × 10−7 m)/(1 m) = 6 × 10−7 rad
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(d) For the radiotelescope
θ = (1.22 × 0.22 m)/(76.2 m) = 3.5 × 10−3 rad
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For best results we want θ small.
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T4
The angle from the axis to the first minimum is given by Equation 5
1. 22 λ
θ =
d
(Eqn 5)
as
θ =
1. 22 λ
= (1.22 × 101cm/ 3001cm)1rad = 0.0407 rad
d
1
The angular spread of the central maximum is therefore twice this angle, i.e. 0.0814 rad.
1
The diameter of this central beam when it reaches the Earth will then be
0.0814 × 2510001km = 2035 km.
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T5
The resolution limit of the normal eye occurs when the two sources are separated by about 1′ of arc i.e. 0.07 mm
at 25 cm.
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(a) If we take the individual letters to be about 1 mm in size, separated by about 1 mm then, by proportion, the
furthest distance of distinguishability is about 3.5 m.
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(b) The Earth–Moon distance from Answer T1 is 3.5 × 108 m and so the minimum distinguishable size is:
3.5 × 108 m × 0.07 mm/(0.25 m) = 98 km.
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T6
The ray diagram of Figure 15b is similar to that for
this system, except that the final image is formed
just beyond the near point. We substitute v =
−30 cm and f = +10 cm in the thin lens equation
(1/v) − (1/u) = 1/f and obtain
1
h'
1
h
1/u = − 1/(30 cm) − 1/(10 cm) = −4/(30 cm),
1
1
1
θI
u
so that
θI
D
u = −7.5 cm.
(b)
1
The transverse magnification
Mtran = h′/h = v/u = −30 cm/(−7.5 cm) = 4.
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Figure 15b Ray diagrams for a magnifying glass with the
object brought closer (i.e. between F and the lens) and the image
formed at the near point. It is assumed that the eye is placed very
close to the lens so that D can be measured from the lens.
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S570 V1.1
T7
In order to maintain the same exposure, when the shutter time is halved, the aperture must be doubled in area
which is achieved by changing to the next smallest f-number. So, if f/11 is the proper stop for (1/125) s, then we
require f/8 for (1/250) s, f/5.6 for (1/500) s and f/4 for (1/1000) s.
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Since the f-number is the focal length f divided by the aperture diameter d, for f/11, d = 50 mm/11 = 4.55 mm;
for f/4, d = 50 mm/4 = 12.5 mm.
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T8
For the final image to be formed at infinity, the intermediate image must be formed at the (first) focal point of
the eyepiece, i.e. 25 mm from the eyepiece. This means that the intermediate image is (180 − 25) mm
i.e. 155 mm from the objective. From the thin lens equation we have 1/u = 1/v − 1/f, so that applying this to the
objective and substituting f = 2 mm and v = 155 mm gives
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1/u = 1/(155 mm) − 1/(2 mm) = (2 − 155)/(310 mm)
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3333221 = −153/(310 mm) and so u = −2.03 mm
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i.e. the object must be placed 2.03 mm in front of the objective.
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T9
For a telescope, for which both the object and final image are effectively at infinity, the length of the tube is
fO + fe = 100 cm. The angular magnification = fO/fe = 40, so fO = 40fe. Substituting for fO
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40fe + fe = 100 cm and fe = 100 cm/41 = 2.44 cm
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Hence fO = (100 − 2.44) cm = 97.56 cm.
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To calculate the exit pupil size we know that the angular magnification is also given by dO/de = 40,
so 12 cm/de = 40 and de = 3 mm.
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∞
T10
erect virtual
image at
infinity
Looking at Figure 28 you
will see how the
introduction of a third
converging lens between
F O and Fe produces an erect
image at infinity which
makes the telescope
suitable for terrestrial use.
Fo Fi
Fe
intermediate
(erecting) lens
objective lens
eyepiece lens
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