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Lehigh University
Physics 21, Spring 2008
April 7, 2008
HW25 Solutions
251.
(HRW 34107)
A fruit fly of height H sits in front of lens 1 on the central axis
through the lens. The lens forms an image of the fly at a distance
from the fly;
the image has the fly's orientation and height
. What are (a) the focal length f
1
of the lens and (b) the object distance p
1
of the fly? The fly then leaves lens 1 and sits in
front of lens 2, which also forms an image at
that has the same orientation as
the fly, but now
. What are (c) f
2
and (d) p
2
?
Solution:
107. (a) In this case
m
> +1 and we know that lens 1 is converging (producing a virtual
image), so that our result for focal length should be positive. Since
P + i
1

= 20 cm and
i
1
= – 2
p
1
, we find
p
1
= 20 cm and
i
1
= – 40 cm. Substituting these
into Eq. 349,
11 1
111
p
if
+
=
leads to
f
1
= +40 cm, which is positive as we expected.
(b) The object distance is
p
1
= 20 cm, as shown in part (a).
(c) In this case 0 <
m
< 1 and we know that lens 2 is diverging (producing a virtual
image), so that our result for focal length should be negative. Since
p + i
2

= 20 cm and
i
2
= – p
2
/2, we find
p
2
= 40 cm and
i
2
= – 20 cm. Substituting these into Eq. 349 leads to
f
2
= – 40 cm, which is negative as we expected.
(d)
p
2
= 40 cm, as shown in part (c).
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 Spring '08
 Kim
 Physics

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