CBSE Class 10 Answered
A screen bearing a real image of magnification m1,formed by a convex lens,is moved by a distance x.The object is then moved until a new image of magnification m2 is formed on screen.The focal lenght of lens is:
(1) . x/m2-m1
(2) . m2-m1/x
(3) . x/m1-m2
(4) . m1-m2/x
Asked by dfojajjar | 16 Feb, 2020, 01:35: PM
We have lens formula :- (1/v) - (1/u) = 1/f ..............(1)
where v is lens-to-image distance, u is lens-to-object distance and f is focal length
By multiplying eqn.(1) with v, we get , 1 - (v/u) = (v/f)
since magnification m of the image is given by, m1 = (v/u),
then we get from above equation, 1 - m1 = (v/f) ...................(2)
if image-to-lens distance is changed to v+x, then magnification becomes m2
Equation (2) with new magnification m2 is written as , 1 - m2 = (v+x)/f = (v/f) + (x/f) ...........(3)
Substituting for (v/f) from eqn.(2) in eqn.(3), we get, 1- m2 = 1 - m1 + (x/f)
Hence we get , (x/f) = [ m1 - m2 ] , hence focal length f = x / [ m1 - m2 ]
Answered by Thiyagarajan K | 16 Feb, 2020, 05:06: PM
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