This preview shows page 1. Sign up to view the full content.
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
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details