An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. find position and nature of the image formed by the lens.
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u=-30cm f= -15cm
1/f=1/v-1/u
1/v=1/f+1/u
on substituting v=-10cm
m=v/u
m=-10/-30
m=1/3=.33
so the magnification is positive so it is a virtual erect image .
m isless than 1 so the image is diminished .
and v is negative so the image is formed on the same side of the lens.
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