A biconvex lens, `f_(1)=20cm` , is placed 5cm in front of a convex mirror, `f_(2)=15cm`. An object of length 2 cm is placed at a distance 10cm from the lens.
Find the loaction and nature of the final image after the leftward travelling rays once again pass through the lens. Find overall magnification.

Answer

Final image is to the right of the lens at a distance 51.1 cm form the lens, real, erect and magnifies, `(8)/(3)`

Explanation

Form lens equation,
`(1)/(v)-(1)/((-10))=(1)/(+20), v=-20cm`
Magnification, `m_(1)=((-20)/(-10))=+2`
Image is virtual, erect and magnified.
The first image acts as an object for the convex mirror. Object distance for the mirror, `u^(')=(20+5)=25cm`
From the equation, `(1)/(v^('))+(1)/((-25))=(1)/(+15), v^(')=+(75)/(8)cm`
Magnification, `m_(2)=((+75//8)/(-25))=(3)/(8)`
Image is virtual (to the right of the mirror), erect and diminished.
The object distance for second refraction at the lens
`=(75)/(8)+5=(115)/(8)`
From lens equation, `(1)/(v^(''))-(1)/((+115//8))=(1)/(-20)`
`v^('')=(460)/(9)=+51.1cm`
Magnification, `m_(3)=((+460//9)/(115//8))=(32)/(9)`
Overall magnification , `m=m_(1)xxm_(2)xxm_(3)` is
`(2)xx((3)/(8))xx((32)/(9))=(8)/(3)`
Hence size of image is `((8)/(3)xx2)cm =5.33cm`
Final image is to the right of the lens at a distance 51.1 cm from the lens, real, erect adn magnified.