A convex lens of focal length 15cm and a concave mirror of focal length 30cm are kept with their optic axes PQ and RS paralledl but separated in vertical direction by 0.6 cm as shown in Figure. The distance between the lens and mirror is 30cm. An upright object Ab of height 1.2 cm is placed on the optics axis PQ of the lens at a distance of 20 cm from the lens. If `A^(')B^(')` is the image after refraction from the lens and reflection from the mirror, find the distance of `A^(')B^(')` from the pole of the mirror and obtain its magnification. Also, locate positions of `A^(') ` and `B^(')` with respect to the optic axis RS.

Answer

15 cm ,`(3)/(2)`

Explanation

(a) Rays coming from object AB first refract from the lens and then reflect from the mirror .
Refraction from the lens :
` u =-20 cm , f = +15 cm `
Using lens formula `(1)/(v) - (1)/(u) = (1)/(f) , (1)/(v) - (1)/(-20) = (1)/(15)`
`therefore v = +60` cm and linear magnification ,
`m_(1) = (v)/(u) = (+60)/(-20) = -3 ` i.e, , first image formed by the lens will be 60 cm from it (or 30 cm from mirror) towards left and 3 times magnified but inverted
Length of first image `A_(1) , B_(1)` would be `1.2 xx 3 = 3.6` cm (inverted).

Reflection from mirror : Image formed by lens `(A_(1) , B_(1))` will behave like a virtual object for mirror at a distance of 30 cm from it as shown . Therefore Using mirror formula ,
`(1)/(v) + (1)/(u) = (1)/(f) "or" (1)/(v) + (1)/(30) = - (1)/(30) therefore v = -15 cm`
& linear magnification , `m_(2) = -(v)/(u) = (-15)/(+30) = +(1)/(2)`
i.e. final image A'B' will be located at a distance of 15 cm from the mirror (towards right) and since magnification is `+^((1)/(2))` length of final image would be `3.6 xx (1)/(2) = 1.8 "cm "`
`therefore A'B' = 1.8` cm
Point `B_(1)` is 0.6 cm above the optic axis of mirror , therefore , its image B' would be `(0.6)(1)/(2) = 0.3` cm
above optic axis . Similarly , point `A_(1)` is 3 cm below the optic axis , therefore , its image A' will be `3 xx (1)/(2) = 1.5 ` cm below the optic axis as shown below :

Total magnification of the image .
`m = m_(1) xx m_(2) = (-3) (+(1)/(2)) = -(3)/(2)`
`therefore A'B' = (m)(AB) = (-(3)/(2))(1.2) = -1.8` cm
Note that , there is no need of drawing the ray diagram if not asked in the questions .