1.

The distance between an object and its image produced by a converging lens is 0.72 m. The magnification is 2. What will be the magnification when the object is moved by 0.04 m towards the lens ?

Answer»

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Solution :(b) Given, `m = 2` "and the IMAGE formed is REAL"
`THEREFORE "For a real image" (1)/(f) = (1)/(v) + (1)/(u)`
`or "" 1 + (1)/(m) = (u)/(f)`
`rArr "" 1+ (1)/(2) = (u)/(f)`
`rArr (3)/(2) = (0.72)/(f)`
`or "" f = (0.72 xx 2)/(3) = 0.16m`
As the object is moved by 0.04 m towards the LENS, the new
`mu_(1) = 0.72 - 0.04 = 0.68 m`
Again real image is formed
`rArr "" 1 + (1)/(m) = (u_(1))/(f)`
`rArr "" (1)/(m) = (0.04)/(0.16)`
`rArr "" m = 4`.


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