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

Options

A. 2
B. 4
C. 3
D. 6

Answer

B

Explanation

Converging lens (or convex lens) forms a real image. Let
`u=-x and v= 2x`
Then `3x= 0.72m`
`x= (0.72)/(3)m= 0.24m = 24cm`
`:. u= -24m and v= 48cm`
According to thin lens formula
`(1)/(v)-(1)/(u) = (1)/(f)`
`:. (1)/(48) + (1)/(24) = (1)/(f)`
`(3)/(48) = (1)/(f) or f= 16cm`
When the object is moved by 0.04m (= 4cm) towards the lens, then new object distance is
`u'= -20cm`
Let v' be new image distance
`:. (1)/(v') -(1)/(-20) = (1)/(16) or (1)/(v') = (1)/(16) - (1)/(20)`
`(1)/(v') = (1)/(80) or v' = 80cm`
+ve sign shows that the image formed by the lens is behind the lens. It is real image.
New magnification, `m' = (v')/(u') = (80 cm)/(-20cm) = -4`
`|m'| =4`