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Solution
ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s30/3/c/solution
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Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford
/
2019
/
S30
/
3
/
c
/
Solution
by
Codex
0
2026-10-25
At maximum
projected separation
, an
Earth
analogue at
10
pc
is separated from its
star
by
ϑ
=
a
/
D
sys
=
0.1
arcsec
=
4.85
×
1
0
−
7
rad
. The
Rayleigh criterion
for
a
circular pupil gives
D
tel
≳
ϑ
1.22
λ
≃
1.26
m
(
0.5
μ
m
)
,
50.3
m
(
20
μ
m
)
.
(1)
The relevant
quantity
is the
star
–
planet
separation, not the tiny
angular diameter
of the
planet
itself. Simple angular
optical resolution
is insufficient: the
planet
lies in the
star
'
s
diffraction
pattern
at reflected
contrast
of order
1
0
−
10
, or thermal
contrast
of order
1
0
−
6
. Residual speckles,
wavefront errors
, scattered
light
, detector/
background noise
and, in the
infrared
, strong
telescope
and
sky
thermal emission obstruct detection.
A
coronagraph
or other stellar suppression method generally has an inner working
angle
of several
λ
/
D
, making practical
aperture
requirements larger than the bare
Rayleigh criterion
. The question'
s
instrumentation
claim is interpreted in its 2019
context
; the physical requirements remain the
basis
of the estimate. Maximum reflected
brightness
near full phase also coincides with small
projected separation
, so maximum
brightness
and maximum resolvable separation cannot generally be obtained simultaneously.
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