A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical directiion by 0.6 cm as shown . The distance between the lens and mirror is 30 cm . An upright object AB of height 1.2 cm is placed on the optic 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 the reflectiion 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 .