Winter 1996
HOMEWORK 4 with Solutions
1.
Find the image of the object for the single concave mirror system shown in Fig.1 (see next pages for
worksheets) by: (a) measuring the radius R and calculating the focal length for the concave mirror, (b)
drawing the focal point into the diagram, and (c) constructing the image of the object graphically with
three rays.
R
R = 10 cm, and f = 2
Three rays:
1
2
3
= 5 cm
Incident Rays
Parallel to the optical axis
Through the center of the mirror (C)
Through the focal point of the mirror (F)
C
F
Object
Reflected Rays
Through the focal point of the mirror (F)
Through the center of the mirror (C)
Parallel to the optical axis
Image
2.
Regarding your own face as a real object, describe the image of your face which you see standing 4 feet
from the center (and looking directly toward) a polished and well-reflecting brass ball 2 feet in diameter
hanging in front of a pawn shop. Determine the image of your face both graphically and by calculation.
By Calculation:
Brass ball is equivalent to a CONVEX MIRROR ( f < 0) .
Diameter = 2 · R = 2 ft 
R = 1 ft and f = – 0.5 ft
You are standing at 4 ft from the center of the brass ball, but the distance s o of object from
the mirror is measured from the vortex of mirror.  so = (4 – 1) = + 3 ft
1
1
1
From the mirror equation, f = s
+ s
o
i
1
1
si = { –0.5 – 3 } -1 = –0.43 ft
Since si < 0, the image is virtural.
yi
si
M = y = – s = + 0.14.
o
o
Since M > 0 and |M| < 1, the image is up-right and minified.
Graphically,
1 ft
Image
so
F
4 ft
C
3.
(A) Calculate the focal length of a thin bi-concave lens (n=1.6) in air, having radii of 6 and 18 cm. (B)
Locate and describe (magnified? inverted? real?) the image of a real object 5 cm from lens. (C) What
changes if the object is virtual instead of real? (D) What changes in (A), (B), and (C) if everything is
immersed in water (n=1.33)?
(A) bi-concave lens  R1 < 0 and R2 > 0 , negative lens.
R1 = – 6 cm, R2 = +18 cm, n = 1.6
By the lens maker's equation,
1
1
1
=
(n
–
1)
{
–
f
R1
R2 }  f = – 7.5 cm
1
1
1
By the lens equation, f = s + s  si = – 3.0 cm
o
i
yi
si
M= y =– s
= + 0.6
o
o
Image is at 3.0 cm in the left side of the lens, virtual ( si < 0 ), up-right ( M > 0 ),
minified ( M < 1).
(B) so = + 5 cm.
(C) For a virtual object, the sign of so is changed from + to –.  so = – 5 cm.
Then, si = + 15.0 cm. M = 3.0
Image is at 15.0 cm in the right side of the lens, real ( si > 0 ), up-right ( M > 0 ),
magnified ( M > 1).
(D) In water (no=1.33), the lens maker's equation is changed to
no
1
1
=
(n
–
n
)
{
–
o
f
R1
R2 }  f = – 22.2 cm
For a real object at 5 cm, si = – 4.1 cm and M = + 0.82
Image is at 4.1 cm in the left side of the lens, virtual ( si < 0 ), up-right ( M > 0 ),
minified ( M < 1).
For a virtual object at 5 cm, si = + 6.5 cm and M = + 1.3.
Image is at 6.5 cm in the right side of the lens, real ( si > 0 ), up-right ( M > 0 ),
magnified ( M > 1).
4.
A positive meniscus thin lens (n=1.48), with radii of curvature 4 and 12 cm, is positioned in contact with
a plano-concave lens (n=1.4) of radius 10 cm. (A) What is the effective focal length and refractive power
of the lens combination? (B) What image (real? magnified? inverted?) will this lens combination produce
from a real object located 5 cm away from it? (C) What will be the focal length of the lens-combination if
the two lenses are separated by D = 15 cm?
Positive meniscus thin lens: R1 = + 4 cm and R2 = +12 cm (or R1=–12cm, R2=–4cm)
1
1
1
=
(n
–
1)
{
–
f
R1
R2 }  f1 = + 12.5 cm
Plano-concave lens: R1 = ∞ and R2 = +10 cm (or R1= –10 cm and R2=∞)
f2 = – 25 cm
(A)
1
1
1
=
+
feff
f1
f2
 feff = + 25 cm
Effective refractive power = Inverse of effective focal length (in meter)
= 4 m-1 = 4 Diopters
(B) so = + 5 cm  si = – 6.25 cm , and M = – si / so = + 1.25
Image is located at 6.25 cm in the left side of the lens combination, virtual (si < 0),
up-right ( M > 0), and magnified ( | M | > 1).
(C) D = 15 cm
f1 ( D – f2)
ff = D – (f + f )
1
2
f2 ( D – f1)
D – (f1 + f2)
fb =
= + 18.2 cm (in the left side of the first lens)
= – 2.27 cm (in the left side of the second lens)
*** Effective focal length
1
1
1
D
=
+
–
 feff = + 11.4 cm
feff
f1
f2
f1 f 2
Then, the principal planes are at L1H1 = – 6.8 cm and L2H2 = – 13.6 cm
10 cm
L1
L2
Fb
Ff
*
F1
F1 F2
H1
feff
H2
feff
*
F2
5.
Two thin lenses (L1 double-convex with focal points F1, and L2 double-concave with focal points F2) are
placed at a distance D as shown in Fig.2 (see next pages for worksheets). Determine the image of object S
using a "two-step process" as discussed in class. Obtain location and magnification of image
(A)
Graphically, using the figure below
(B)
By numerical calculation using thin lens equations for L1 and L2
If done right (with proper signs) results from (A) and (B) should agree. You may compare your results
with the one obtained using the equation for si2 derived and formulated on page 5 of Handout I. (Use the
appended worksheet to find the graphical solution.)
(A) Graphically,
L1
L2
Image
of L2
S
Image of L1
Object of L2
F1
F1F2
F2
D
(B) By calculation, (All parameters are measured by a ruler.)
f1 = + 3.0 cm, f2 = – 2.5 cm, D = 6.0 cm, so1 = + 4.0 cm, yo1 = 1 cm
For the lens L1, si1 = + 12.0 cm and yi1 = – 3 cm.
 Image of L1 is real, inverted, and magnified.
For the lens L2, so2 = D – si1 = – 6 cm and yo2 = yi1 = – 3 cm.
 si2 = – 4.3 cm and yi2 = + 2.15 cm
 Image of L2 is virtual, up-right, and magnified (compared to the S).

Good agreement with Graphical solution!
** From the page 5 of Handout I ,
f2 D (so1 – f1) – f1 f2 so1
si2 = (D – f )(s – f ) – s f
2 o1
1
o1 1
= – 4.3 cm (Good agreement!)
Image
(A)
Air
Air
C2
C1
Object
F
F
n = 1.65
(B)
Image
Water n=1.33
Water
F
C1
C2
F
n=1.7
Image
(C)
Air
Air
F
Virtual
Object
C1
n=1.4
F
7*. Optional
Find graphically the image of the object for two plane mirror system shown in Fig.4 (see next pages for
worksheets) where two mirrors are inclined to each other at 90°. Draw all possible images and two rays
(at least) for each image.
Object
90°