(1) d0 = 20 cm, di = 38 cm, hi = -40 cm
1
f
=
1
do
+
1
di
f = + 13.1 cm
M = – di
do
=–
1
1
1
=
+
f
20
38
converging
inverted,real
lens
38
20
= - 1.9
M(ho) = hi
- 1.9(ho) = - 40
ho = 21 cm
object was originally
21 cm tall
(2)
(3)
real, inverted, smaller,
between f’ and 2f’
virtual, upright, larger,
same side of the lens
(4)
virtual, upright, smaller,
same side of the lens
(5)
Given:
M = +1/5
f = - 2.4 cm
do = ?
M is positive because
the image is virtual
f is negative because
the lens is diverging
f
1
M = - di/do
1/5 = -di/do
1
-2.4
Cross multiply
do = - 5di
1
-2.4
 do = 9.6 cm
1
=
- 4 -2.4
- 5di
=
1
do
1
=
-5di
=
+
+
1
-5di
+
1
di
1
di
-5
-5di
di = -1.92 cm
(6) Given:
do = 20 cm
f = 18 cm
1
1
=
+
f
do
1
1
di
18
20
+
1
di
Real,Inverted,
larger
di = +180 cm
180
d
i
M=–
=–
20
do
=
1
= -9
Image is 180 cm
to the right of the
lens
Image is real,
inverted, and
larger