10. (a)
f
= +20 cm (positive, because the mirror is concave);
r
=2
f
=2
(+
2
0cm
)=+
4
0cm
;
i
=
(1
/f
−
1
/p
)
−
1
=(1
/
20 cm
−
1
/
10 cm)
−
1
=
−
20 cm;
m
=
−
i/p
=
−
(
−
20 cm
/
10 cm) = +2
.
0. The
image is virtual and upright. The ray diagram would be similar to Fig. 358(a) in the textbook.
(b) The fact that the magni±cation is 1 and the image is virtual means that the mirror is ﬂat (plane).
Flat mirrors (and ﬂat “lenses” such as a window pane) have
f
=
∞
(or
f
=
−∞
since the sign
does not matter in this extreme case), and consequently
r
=
∞
(or
r
=
−∞
) by Eq. 353. Eq. 354
readily yields
i
=
−
10 cm. The magni±cation being positive implies the image is upright; the answer
is “no” (it’s not inverted). The ray diagram would be similar to Fig. 356(a) in the textbook.
(c) Since
f>
0, the mirror is concave. Using Eq. 353, we obtain
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 Fall '08
 SPRUNGER
 Physics

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