A diverging lens has a focal length of magnitude 24.8 cm. (a) Locate the images
ID: 1731998 • Letter: A
Question
A diverging lens has a focal length of magnitude 24.8 cm.
(a) Locate the images for each of the following object distances.
49.6 cm
---Select---
in front of the lens
behind the lens
24.8 cm
---Select---
in front of the lens
behind the lens
12.4 cm
---Select---
in front of the lens
behind the lens
(b) Is the image for the object at distance 49.6 real or virtual?
real
virtual
Is the image for the object at distance 24.8 real or virtual?
real
virtual
Is the image for the object at distance 12.4 real or virtual?
real
virtual
(c) Is the image for the object at distance 49.6 upright or inverted?
upright
inverted
Is the image for the object at distance 24.8 upright or inverted?
uprightinverted
Is the image for the object at distance 12.4 upright or inverted?
uprigh
tinverted
(d) Find the magnification for the object at distance 49.6 cm.
Find the magnification for the object at distance 24.8 cm.
Find the magnification for the object at distance 12.4 cm.
---Select---
in front of the lens
behind the lens
Explanation / Answer
a)
f = focal length of the diverging lens = - 24.8 cm
do = object distance = 49.6 cm
di = image distance = ?
using the lens equation
1/do + 1/di = 1/f
1/49.6 + 1/di = 1/(- 24.8)
di = - 16.5 cm
in front of the lens
f = focal length of the diverging lens = - 24.8 cm
do = object distance = 24.8 cm
di = image distance = ?
using the lens equation
1/do + 1/di = 1/f
1/24.8 + 1/di = 1/(- 24.8)
di = - 12.4 cm
in front of the lens
f = focal length of the diverging lens = - 24.8 cm
do = object distance = 12.4 cm
di = image distance = ?
using the lens equation
1/do + 1/di = 1/f
1/12.4 + 1/di = 1/(- 24.8)
di = - 8.27 cm
in front of the lens
b)
Virtual
Virtual
Virtual
c)
Upright
Upright
Upright
d)
m = - di/do = - (- 16.5)/49.6 = 0.33
m = - di/do = - (- 12.4)/49.6 = 0.25
m = - di/do = - (- 8.27)/49.6 = 0.17
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