CBSE Class 10 Answered
An object is placed on the principle axis of a concave lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find nature and position of the image
Asked by akshaypolishetty | 06 Jan, 2021, 03:18: PM
Expert Answer
We have lens equation :- (1/v) - (1/u) = 1/f
where v = lens-to-image distance
u = lens-to-object distance = -15 cm
f = focal length of concave lens = -10 cm
( Cartesian sign convention is followed)
Hence , we get , ( 1/v ) + ( 1/10 ) = ( -1/15 )
1/v = (-1/15) -(1/10) = -1/6
Hence v = - 6 cm
image is formed at a distance 6 cm from lens on the same side of object
Magnification m = v / u = (-6) / (-15 ) = 0.4
Image is diminished , erect and virtual
Schematic of ray diagram is given below
Answered by Thiyagarajan K | 06 Jan, 2021, 03:44: PM
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