24.30. Set Up:
Since the image is inverted,
and
Solve:
gives
gives
and
The object is 154 cm to the left of the lens. The image is
217 cm to the right of the lens and is real.
24.31. Set Up:
Since the image is projected onto a wall, the image is real and
and
both
s
and
are positive, so
The height of the image is three times the height of the object, so
and
Solve: (a)
so
You are 0.60 m from the lens.
(b)
so the image is inverted.
(c)
and the lens is converging.
Reflect:
Since the image is real the lens must be converging and your distance from the lens must be greater than the
focal length.
24.32. Set Up:
The type of lens determines the sign of
The sign of
depends on
whether the image is real or virtual.
is positive because the image is on the side of the lens
opposite to the object.
Solve: (a)
and
is positive so the lens is converging.
(b)
The image is 1.88 mm
tall.
and the image is real.
24.33. Set Up:
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.
 Spring '07
 Shoberg

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