Page 407 - Schaum's Outline of Theory and Problems of Applied Physics

392 SPHERICAL MIRRORS [CHAP. 31

384,000 km from the earth. (a) How far should the photographic plate be placed from the mirror? (b) What
will be the size and nature of the moon’s image?

(a) The object distance is so much greater than the mirror’s focal length of f = R/2 = 2 m that we can let p =∞
here. Substituting 1/p = 0 and f = 2 m into the mirror equation yields
1 1 1 1 1
+ = 0 + = q = 2m
p q f q 2m
The image is real and on the same side of the mirror as the moon.
8
6
(b) Sincethemoon’sdiameteris h = 3500km= 3.5×10 mandobjectdistanceis p = 384,000km= 3.84×10 m,
the diameter of the moon’s image is
q 2m
6
h =−h =−(3.5 × 10 m) =−0.018 m =−18 mm

p 3.84 × 10 m
8
The minus sign indicates an inverted image (Fig. 31-10).
In general, an object very far away from a concave mirror relative to its focal length will have a real,
inverted image smaller than the object and located very nearly at the focal point of the mirror.

Fig. 31-10 Fig. 31-11

SOLVED PROBLEM 31.8
A concave mirror has a radius of curvature of 120 cm. How far from the mirror should one’s face be so
that the image is erect and twice the size of the actual face? Is the image real or virtual?

The focal length of the mirror is f = R/2 = 60 cm and is positive since the mirror is concave. Because the
image is to be erect and twice the size of the object, m =+2. We know m and f and are to find the object distance p.
There are various ways to do this, one of which is as follows. Since m =−q/p,wehave q =−mp. The negative
image distance signifies a virtual image (Fig. 31-11). Substituting q =−mp in the mirror equation enables us to
solve for p:
1 1 1 1 1 1 m − 1 1 mp
+ = − = = = f
p q f p mp f mp f m − 1
f 60 cm
p = (m − 1) = (2 − 1) = 30 cm
m 2
In general, an object placed closer to a concave mirror than its focal point (that is, closer than f ) will have a
virtual, erect image that is larger than the object.

SOLVED PROBLEM 31.9
How far away from a concave mirror of 40-cm focal length should a real object 30 mm long be located
in order that its image be 8 mm long?

When a concave mirror forms an image that is smaller than the object, the image is always inverted (see
Fig. 31-9). An inverted image is considered to have a negative height, so here the magnification is
h −8mm
m = = =−0.267
h 30 mm