A diverging lens has a focal length of magnitude 21.6 cm. (a) Locate the images
ID: 1454569 • Letter: A
Question
A diverging lens has a focal length of magnitude 21.6 cm.
(a) Locate the images for each of the following object distances.
43.2 cm
distance
cm
location
---Select---
in front of the lens
behind the lens
21.6 cm
distance
cm
location
---Select---
in front of the lens
behind the lens
10.8 cm
distance
cm
location
---Select---
in front of the lens
behind the lens
(b) Is the image for the object at distance 43.2 real or virtual?
real
virtual
Is the image for the object at distance 21.6 real or virtual?
real
virtual
Is the image for the object at distance 10.8 real or virtual?
real
virtual
(c) Is the image for the object at distance 43.2 upright or inverted?
upright
inverted
Is the image for the object at distance 21.6 upright or inverted?
upright
inverted
Is the image for the object at distance 10.8 upright or inverted?
upright
inverted
(d) Find the magnification for the object at distance 43.2 cm.
Find the magnification for the object at distance 21.6 cm.
Find the magnification for the object at distance 10.8 cm.
distance
cm
location
---Select---
in front of the lens
behind the lens
Explanation / Answer
f = - 21.6 cm
(a)
1)
do = 43.2 cm
We know,
1/f = 1/di + 1/do
-1/21.6 = 1/di + 1/43.2
di = -14.4 cm
Location = Infront of the Lens.
2)
do = 21.6 cm
We know,
1/f = 1/di + 1/do
-1/21.6 = 1/di + 1/21.6
di = -10.8 cm
Location = Infront of the Lens.
3)
do = 10.8 cm
We know,
1/f = 1/di + 1/do
-1/21.6 = 1/di + 1/10.8
di = -7.2 cm
Location = Infront of the Lens.
(b)
Diverging lens allways forms Virtual Images.
All the Images are Virtual.
(c)
Diverging lens allways forms Upright Images.
All the Images are Upright.
(d)
Magnification is given by,
M = -di/do
(1)
M = 14.4/43.2 = 0.33
(2)
M = 10.8/21.6 = 0.5
(3)
M = 7.2/10.8 = 0.66
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