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Unformatted text preview: 36.2.1 Convex Mirror s E 36 .2 .2 Convex Mirror Based on the ray tracing approach determine the properties ofthe l IRI
image for the setup: The object is to the left ofthe convex mirror. p \ Consider a convex mirror with radius |R|:1m. The object
distance p=1m. Locate the image distance q. Determine
virtual the magnification of the image. |M|>1 |M|<1 Hint: Ray tracing rules:
0 When the ray hits the symmetry point S, it is reﬂected
symmetrically. Hint: 1/p+1:’q = 2/R = 1/f, R=4_-|R|. M=h’/h= - q/p. For
0 When the ray travels along the radial direction, upon hitting the M>0 the image is upright. For M<0, the image is inverted. minor, its direction is reversed. I . .
Explanation: For the convex mirror, R being the Extra: Consider abundle of parallel rays near the horizontal axis coordinate of center of the spherical surface, it is negative,
hitting the mirror. Locate the point where the extrapolation ofthe i-e- R=-IBI_- 1/f1=(2/R) - Up = -(2/IRI)-1/p = -3, q=-1/3. So
reﬂected rays will converge? the mgmflcatlon: M: -q-’p = 1/3. Ans=3. Explanation: From the sketch below one sees that the image must
be reduced and it is virtual. Ans=3. Extra: The reader should check by the ray-tracing method that the
convergent point is located at ahalfway between S and 0. 36.4.1 A thin lens 36.3.1 Convergent or divergent? Consider the setup shown above. Given: The object distance p=f/2. Consider a spherical surface AOB, where nL>nR. Is it a Find: The image dismnce Cl- ' 9 . .
convergent oradwergem surface" Extra: Based on the ray traclng approach, determine Chooge one; whether the image is virtual or real? Converent surface Choose one- l
q f/2 Neither of the above Hint: Consider how a parallel ray would be refracted.
Hint: l/p + 1/q= l/f.
Explanation: Inspection on the ray diagram shows
9R>0L- This Show that ‘11? surface is convergent Explanation: 1/q=1/f —1/p= —1/f. So q= —f. A negative
Ans=1- image distance implies that it is virtual. See sketch. Ans=4. Extra: From the ray diagram here, one sees that the image
is virtual. 36.4.2 Lens maker’s formula A
B Find the focal length of the thin lens above, Which has a
crescent shape and it is facing to the left. Here |RA|=a, nlens=1 and 1’lmedz1 . l 2 3 4
f 4a 2a —2a -43 Hint: The lens maker’s formula is given by:
1/f= (nlmS/n -1)(1/R1-1/R2). Here R1 is the medium coordinate of the center of the ﬁrst spherical surface With
the origin being at the lens, and R2, that of the second
surface. Extra: What is f, if the lens is ﬂipped over with the
crescent facing to the right? Explanation: l/f = (1.5/1 - 1)[-1/|RB| + 1/ |RA|] =
0.5[-1/(2a) + Ila] = 0.25/a, This leads to f=4a. Ans=1. Extra: If ﬂipped over, 1/f= (1.5/1 — 1)[l/|RA| - l/ |RB|] =
0.5[1/a - 1/(2a)] = 0.25/a. Thus the focal length remains
the same. 36.4.4 Curvature in 2-lens problem #1 #2 . m to» f
f <1> Y‘kb <3>" \<4> f P1 d=1.5f
Given two identical thin lenses, f1=f2=f The object is located at distance p1=2f from lens #1. The distance of separation d=1.5f. Determine the curvature C3 before the
passage through lens #2. 1 2 3 4
C -2/f -l/f l/f 2/f 3 Hint: Curvature C=J_r1/|R|. Here C2= C1+l/f=1/(2f). In the last step, Cl: -1/p1= -1/(2f) was used.
Extra: Show the curvature at <4>, i.e. immediately after the passage through lens #2 is 3f, or q2=0.33f. Explanation: From the sketch, C3=l/(q1-d)=l/(0.Sf)=2/f.
Extra: C4: C3+1/f= 3/f=1/q2. q2=0.33f. 36 .4 .3 Two—lens problem
#1 #2 p1 d
Given two identical lenses, f1=fz=f. The object is located
at a distance p1=2f from lens #1. The separation is d=5f. Based on the ray diagram determine the property of the
imae formed b the two lens s tern. Hint: Lens #1 alone gives a real image at ql=p1:2f.
Extra: Show the final image is to the right of lens #2 at a
distance q2=1.5f. Explanation: We indicate the situation qualitatively, and
leave the contruction of an accurate diagram to the reader.
The image formed by lens #1 is inverted & real. It
becomes the object of #2. The image formed by #2 is
again real & inverted. Put all together, the final image is
upright & real. Extra: The object distance to the lens #2, p2=d—q1=3f. For
#2, this gives 1lq2=1/f—1/p2=1/f—1/(3f)=2/(3f). So the final
image is to the right of lens #2, at q2=1.5f. #1 #2 36 .4 .5 Two-lens problem Given: Focal lengths of #1 & #2 are: flza, & f2: — a. The separation is d=1.5a, and the object is at p1:2a from
#1. Find: Find the location of the “object” for lens #2, p2.
1 2 3 4 | P2 I a 0.5a —0.5a —a |
Hint: Check first q1=2a. Notice that the image of #1 is
the “virtual object” for lens #2.
Extra: Show that qzza. Explanation: Using 1/q1= —1/p1+1/f= —1/(2a)+1/a:1/(2a),
one gets q1=2a. This leads to pzzd-qlzlﬁa - 2a 2 — 0.5a.
Extra: 1/q2=—1/p2+1/f2= —1/(—0.5a)+1/(—a). So q2=a. Digression: The method of curvature may help one to
visualize the situation. For lens #1 with C1=—1/(2f), C2 = C1+1/f=1/ (2f). This implies that the curvature before
#2 is C3: 1/(q1—d)=1/(0.5a). So the curvature immediately
behind #2, C4=C3+llf2 = 1/(0.5a) - 11a 2 Ila. This leads to q2=1/ qua. ( The sketch is not drawn to
scale.) #1 #2
I <1} <2> x
<3} /' <4> ...
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This note was uploaded on 02/04/2012 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
- Fall '08