A converging lens has a focal length of 93.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.)
(a) 93.0 cm
| q | = cm |
| M | = |
Select all that apply to part (a).
real
virtual
upright
inverted
no image
(b) 11.6 cm
| q | = cm |
| M | = |
Select all that apply to part (b).
real
virtual
upright
inverted
no image
(c) 837 cm
| q | = cm |
| M | = |
Select all that apply to part (c).
real
virtual
upright
inverted
no image
Focal length of lens (f) = 93 cm
(a) distance of the object (do) =93 cm
hence the object is at the focus therefore the image will be at
infinity
so q = 0 , M= 0
and no image will be seen
(b) distance of the object (do) = 11.6 cm
from the lens formula
1/do + 1/q = 1/f
where q is image distance
1/q = 1/f - 1/do
1/q = (1/93 )- (1/11.6)
q = -13.25 cm
Magnification (M) = -(q/do) = -(-13.25/11.6)
M = 1.14
hence the image will be virtual and upright
(c)
distance of the object (do) = 837 cm
1/do + 1/q = 1/f
1/q = 1/f - 1/do
1/q = (1/93 )- (1/837)
q = 104.625 cm
M = - (q/do) = -(104.625/837)
M = -0.125
therfore the image will be real and inverted.
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