A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When liquid of refractive index is put between the lens and the mirror , the oin has to moved to A', such that OA'=27cm, to get its inverted real imaged at A' itself. The value of `mu_(l)` will be

Options

A. `(4)/(3)`
B. ` sqrt(2)`
C. `sqrt(3)`
D. ` (3)/(2)`

Answer

A

Explanation

Initially image is formed at A itself
`:.` After refraction from lens the rays must be incident normally on the plane mirror
`(1)/( v) - (1)/( -OA) = (1)/(f)`
`f = OA = 18 cm `
`(1)/(f) = ((3)/( 2) -1) ((1)/(R) - (1)/(R))`
R = f = 18 cm
After filling the liquid between lens and mirror and placing the object at A. same thing occurs
`(mu_(1))/(v) = - (1)/(OA) = ((3)-1)/(R) + (( mu_(1) + (3)/(2))/(-R))`
`rArr (mu_(1))/( oo) + (1)/(27) = (1)/(36) - (mu_(1) + (3)/(2))/(18) rArr (mu_(1) - (3)/(2))/( 18) = (1)/(36) - (1)/(27) `
`rArr (mu_(1) - (3)/(2))/(18) = (-9)/(36 xx 27) rArr mu_(1) = (3)/(2) - (1)/(6) = (4)/(3)`