Phys 197 Homework Solution 34B
P71.
What is the size of the smallest vertical plane mirror in which a woman of height h can
see her full-length image?
————–
Half of her height, or h/2.
Sketch the person with the mirror to the right. Consider the ray leaving the foot that reflects
and reaches the eye. The reflection point marks the lowest point that is required, and this
point is at a height halfway between the floor and eye. Similar considerations apply to the ray
leaving the top of the hair-do. Real mirrors are longer so as to avoid adjusting the height at
which the mirror is hung, and also to include background in the reflection.
P74.
Rear-View Mirror. A mirror on the passenger side of your car is convex and has a
radius of curvature with magnitude 18.0 cm. (a) Another car is behind your car, 9.00 m from the
mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 m tall, what is the
height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than
they appear. Why is this so?
————–
(a) First, the mirror’s focal length is given by
1/f = 1/2R ⇒ f = R/2 = (−18 cm)/2 = −9 cm = −0.09 m. The radius is negative
because the center of curvature is opposite the outgoing light. In any case, the focal length of
a convex mirror is negative.
We will need s′ :
s′ = f s/(s − f ) = (−0.09 m)(9 m)/(9 m − (−0.09 m)) = −0.089 m.
m = −s′ /s ⇒ y ′ = −ys′ /s = −(1.5 m)(−0.089 m)/(9 m) = 0.0148 m.
(b) We have to consider the angular size as seen by the driver’s eye (more important than what
the passenger sees), and compare it to the angular size for a flat mirror. Let’s take the distance
from the driver’s eye to the mirror as 1.3 m.
Then θ, the angular size in a flat mirror, would be;
θ = (1.5 m)/(1.3 m + 9 m) = 0.146 rad.
And θ′ , the angular size in the curved mirror, would be:
θ′ = (0.0148 m)/(1.3 m + 0.089 m) = 0.0107 rad.
Evidently, the angular size is 15 X smaller in the convex mirror. For the passenger perspective,
try 0.3 m instead of 1.3 m.
P91.
An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the
lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to
refocus the image. Determine the focal length of the lens.
————–
Sketching the first situation, we see that s1 is unknown, and s′1 = +30 cm; since the image is
formed on the screen, it is evidently a real image and the focal length will therefore be positive,
and s′1 is positive.
For the second situation, s2 = s1 + 4 cm and s′2 = 22 cm. (N.B. 22!)
The focal length is given f = ss′ /(s + s′ ). Plug this equation (with values in cm, suppressing
units, and using the symbol s1 for the unknown value) with what is known to get:
30 s1 /(s1 + 30) = (s1 + 4)(22)/(s1 + 4 + 22). Noting that 4 + 22 = 26, cross-multiply to get
30(s21 + 26s1 ) = 22(s21 + 34s1 + 120). Simplify to
s21 + 4s1 − 330 = 0. Solve to get s1 = 16.28 cm. Now that s1 is known, plug back into the
equation for f to get
f = 10.55 cm.
P108.
Angular Magnification. In deriving Eq. (34.22) for the angular magnification of
a magnifier, we assumed that the object is placed at the focal point of the magnifier so that the
virtual image is formed at infinity. Suppose instead that the object is placed so that the virtual image
appears at an average viewer’s near point of 25 cm, the closest point at which the viewer can bring
an object into focus. (a) Where should the object be placed to achieve this? Give your answer in
terms of the magnifier focal length f . (b) What angle will an object of height subtend at the position
found in part (a)? (c) Find the angular magnification M with the object at the position found in
part (a). The angle θ is the same as in Fig. 34.51a, since it refers to viewing the object without the
magnifier. (d) For a convex lens with f = +10.0 cm, what is the value of M with the object at the
position found in part (a)? How many times greater is M in this case than in the case where the
image is formed at infinity? (e) In the description of a compound microscope in Section 34.8, it is
stated that in a properly designed instrument, the real image formed by the objective lies just inside
the first focal point F1′ of the eyepiece. What advantages are gained by having the image formed by
the objective be just inside F1′ as opposed to precisely at F1′ ? What happens if the image formed by
the objective is just outside F1′ ?
————–
I hope you took good notes in class. Units will be cm, and I will frequently suppress them.
(a) Since the image is where the light doesn’t go, s′ is negative, and its magnitude is the
standard near point of 25 cm So,
s′ = −25 cm and s = f s′ /(s′ − f ) = f (−25 cm)/(−25 cm − f ) = f (25/(f + 25)).
(b) Divide the equation for s by s′ to get
s/s′ = f /(s′ − f ). Invert and multiply by −1 to get
−s′ /s = (f + 25)/f . The value s′ = −25 cm has been substituted on the RHS. The LHS is
the lateral magnification m. Since this image is formed at a distance of 25 cm, its subtended
angle θ′ is:
θ′ = ym/25 = y[(f + 25)/f ]/25.
(c) M = m, since the image is at the standard near point distance.
M = (f + 25)/f = 1 + 25/f . In class, I had 1/β for the first term. The answer to this part
is just what happens when β, which must be greater than or equal to one, reaches its limiting
value of 1.
(d) M = 1 + 2.5 = 3.5 X, With the image at infinity, M0 = 25/f = 2.5 X. The ratio is evidently
3.5/2.5 = 1.4.
(e) By having the lens in your eye work, you get more magnification. If the image is outside
F1′ , it will be out of focus, except to a far-sighted person.