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CHAPTER 36 IMAGE FORMATION^ AND^ OPTICAL
INSTRUMENTS
ActivPhysics can help with these problems:
All activities in Section 15
Sections 36-1 and 36-2: Plane and Curved
Mirrors Problem
- A shoe store uses small floor-level mirrors@ let customers view prospective purchases. At what angle should such a mirror be inclined so that a person standing 50 cm from the mirror with eyes 140 cm off the floor can see her feet?
Solution A small mirror (M) on the floor intercepts rays
coming from a customer's shoes (O), which are
traveling nearly parallel to the floor. The angle to the customer's eye (E) from the mirror is twice the angle of reflection, so tan 2a = h/d, or a = tan-' x (140150) = 35.2", for the given distances. Therefore, the plane of the mirror should be tilted by 35.2" from the vertical to provide the customer with a floor-level view of her shoes.
Problem 1 Solution.
Problem
- Two plane mirrors occupy the first four meters of the positive a and y-axes, as shown in Fig. 36-44.
Find the locations of all images of an object a t z = 2 m , y = l m.
FIGURE 36-44 Problem 2 Solution.
Solution In addition ,to the two direct images in each mirror (one reflection) at I, = (-2 m, 1 m) and I, = (2 m, -1 m), a multiple image (two reflections) also appears a t I,, = (-2 m, -1 m). (The latter is the image in one mirror of the direct image in the other.) Ray tracing confirms this, as shown on Fig. 36-44. No more than two reflections are possible for perpendicular mirror planes, so these are all the images.
Problem
- (a) What is the focal length of a concave mirror if an object placed 50 cm in front of the mirror has a real image 75 cm from the mirror? (b) Where and what type will the image be if the object is moved t o a point 20 cm from the mirror?
Solution (a) The mirror equation relates the given distances (both positive for a real object and image) to the focal length: f-' = (50 cm)-I + (75 cm)-l, or f = 30 cm. (See Equation 36-2.) (b) A second application of the
mirror equation yields (el)-' = (30 cm)-' - (20 cm)-',
or P = -60 cm. A negative distance indicates a
virtual, erect image located behind the mirror. (Fig. 3 6 8 c and Table 36-1 confirm these results.)
Problem
- A virtual image is located 40 cm behind a concave mirror with focal length 18 cm. (a) Where is the object? (b) By how much is the image magnified?
Solution (a) The mirror equation (Equation 36-2) gives e = fr/(P1 - f ) = (18 cm)(-40 cm)/(-58 cm) = 12.4 cm (positive distances are in front of the mirror, negative distances behind). (b) Equation 36- 1 gives M = -el/t? = +40 cm112.4 cm = 3.22.
Problem.
- An object's image in a 27-cm-focal-length concave mirror is upright and magnified by a factor of 3. Where is the object?
Solution An upright image in a concave mirror must be virtual, so M = +3 = -el/!. The mirror equation gives ( i / e ) + ( i / r ) = ( l i e ) - (ij3e) = llf,or e = (g)f = ($)(27 cm) = 18 cm (positive in front of the mirror).
Problem
- You look into a reflecting sphere 80 cm in diameter and see an image of your face a t one- third its normal size (Fig. 36-46).. How far are you from the sphere's surface?
Solution The sphere reflects like a convex mirror of focal length
f = R/2 = -40 cm/2 = -20 cm. (Only a convex
mirror produces a reduced, upright, virtual image.) The equation in the solution to Problem 5(a) can be used to find the object distance in terms off and the magnification, t? = f (1 - h/hf) = f (1 - 1/M) = -20 cm(1- 3) = 40 cm.
Section 36-3: Lenses
Problem
17. A light bulb is 56 cm from a convex lens, and its
image appears on a screen located 31 cm on the
. - -^ other^ side of the lens. (a) What is the^ f m d^ len@?!
of the lens? (b) By how much is the image enlarged or reduced?
Solution (a) The object and image distances are both positive, for a real image formed by a single lens (recall that only a real image can appear on a screen), so the lens equation gives f-I = (56 cm)-' + (31 cm)-' = (20.0 cm)-'. (b) Equation 36-4 gives a magnification of M = -31156 = -0.554, so the inverted image is reduced to nearly 55% of the actual size of the bulb.
- A simple camera uses a single converging lens to focus an image on its film. If the focal length of the lens is 45 mm, what should be the lens-to-film distance for the camera to focus on an object 80 cm from the lens?
Set L = 80 cm and f = 45 mm in the lens equation
and solve for t'. The result is t?' = ef / ( t? - f ) = (80x45 cm)/(800 - 45) = 4.77 cm.
problem
27. A lens has focal length f = 35 cm. Find the type and height of the image produced when a 2.2-cm-high object is placed at distances (a) f + 10 cm and (b) f - 10 cm.
Solution The lens equation and magnification for a thin (converging, i.e., positive f ) lens, Equations 3 6 4 and 5, give M = -el/! = - f/(! - f), SO h1 = Mh = - j h + (P- f ). (a) I f f = 3 5 cm a n d e = j + 1 0 c m , then
- h'-^ =^ -(35^ cm)(2.2/10)^ =^ -7.7^ em.^ A^ negative image. -- height signifies a real, inverted image. (b) If e = f - 10 cm, then h' = -(35 cm)(2.2)/(-10) = +7.7 cm, which represents a virtual, erect image of the same size.
