Answer:
(a) As we have
i.e.
This image
formed by convex lens appearsas an object placed at 30 - 8 cm = 22 cmfrom the
convex lens on the right hand side.
Using
we
have,
i.e.
The
parallel beam of light appears todiverge from (220-8) ==212 cm from theconvex
lens from the left.
Now let the
parallel beam of light fall on theconcave lens after''which the convex lens
isplaced at 8 cm.
Then
or
the
image is formed -20 cm to the left ofconcave length and .'. it is 20 + 8 = 28
cmto the left of convex lens.
Again,
i.e.
i.e.
the
light appears to diverge from a pointat 420 cm to the left of convex lens.
Clearly the
answer depends upon the side ofincidence so the notion of effective focallength
is not useful in this case.
(b) Using we
have
i.e.
Magnification
produced by convex lens,
Now (i.e.
objectis to the right of concave lens)
or
magnification
produced by concave lens,
Net
magnification
Size
of image = m x size of object
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