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

`A'b'` at`15`cm to the right of mirror.`B'is`0.3`cm above`RS`and `A'is`1.5`cm below RS.Magnification is`1.5`

Explanation

Image due to lens
`(1)/(v)-(1)/(u)=(1)/(f) rArr (1)/(v_(1))-(1)/(-20)=(1)/(15) rArr (1)/(v_(1))=(1)/(15)-(1)/(20)`
`(1)/(v_(1))=(4-3)/(60) rArr v_(1)=60cm`
`(A_(1)B_(1))/(AB)=(v)/(u) rArr A_(1)B_(1)=(60)/(-20)xx1.2 rArr A_(1)B_(1)=-3.6cm`
image due to Mirror image`A_(1)B_(1)` acts as object for mirror
`(1)/(v)+(1)/(u)=(1)/(f) rArr (1)/(v)+(1)/(+30)=(1)/(-30) rArr v=-15cm`
` m=-(v)/(u) rArr m=-(-15)/(30) rArr m=(1)/(2)`
`m=(A'B')/(A_(1)B_(1)) rArr A'B'=(A_(1)B_(1))/(2) =-1.8cm`
Distance of `A'B'` from pole of mirror =`15 cm` towards right.Net magnetication=`(A'B')/(AB)=(-1.8)/(1.2)=-1.5`
A' while be` (-3)/(2) =-1.5cm,1.5cm` below the`RS`,B' will be `(0.6)/(2)=0.3cm,0.3cm` above `RS.`