An equiconvex lens, `f_(1)=10cm`, is placed 40 cm in front of a concave mirror, `f_(2)=7.50cm` as shown in Figure ., An object 2 cm high is placed 20cm to the left of the lens . Find the position of the final image is formed when leftward travelling rays once again pass through the lens. Find overall magnification.

Answer

Final image is real, inverted and lies 15.6cm to the left of the lens with overall magnification 0.333

Explanation

From lens equation,
`(1)/(v)-(1)/((-20))=(1)/(10),v=-20cm`
Magnification, `m_(1)=(v)/(u)=((+20)/(-20))=-1`
Image is real and inverted, same size as object.
The first image acts as object for concave mirror. Object distance for mirror is `(40-20)` cm.
From mirror equation,
`(1)/(v^('))+(1)/((-20))=(1)/((-7.5)), v^(')=-12cm`
Magnification, `m_(2)=((u^('))/(v^(')))=-((-12)/(-20))=-0.6`
The second image is 12cm to the left of th emirror, real, erect (that is reinverted).
The second image acts as object for the lens. The object distance for second refraction at the lens, `u^('')=+28cm`
From lens equation, `(1)/(v^(''))-(1)/((+28))=(1)/(-10), v^('')=-15.6 cm`
Note the sign convention for f and u.
Magnification, `m_(3)=(v^(''))/(u^(''))=((-15.6)/(+28))=-0.556`
Final image is real, inverted and lies 15.6 cm to the left fo the lens.
Overall magnification,
`m=m_(1)xxm_(2)xxm_(3)`
`=(-1)xx(0.6)xx(-0.556)=-0.333`