Here, `f = 20 cm`. The farhter face of the convex lens of `R_(2) = 22 cm` is silvered. When a ray of light from object `AB | u = -10 cm |` Falls on the silvered lens, first of all, its image is formed due to refraction by convex lens, then due to reflection by the silvered surface acting as a concave mirror `(f_(m) = R//2)` and finally again due to the refraction through the lens. This is because the rays reflected from the silvered surface are refracted again by the lens.
If `F` is equivalent focal length of the system (lens concave mirror and lens) then
`(1)/(F) = (1)/(f) + (1)/(f_(m)) + (1)/(f)`
`(1)/(F) = (1)/(20) + (1)/(11) + (1)/(20) = (11 + 20 + 11)/(20 xx 11) = (42)/(20 xx 11)`
`F = (20 xx 11)/(42) = (110)/(21)`
As the system reflects back and converges the rays, it acts as a concave mirror of focal length
`F = (110)/(21)cm`.
As `(1)/(v) = (1)/(F) - (1)/(u), and u = - 10cm, F = -(110)/(21)cm`
`:. (1)/(v) = -(21)/(110) - (1)/(-10) = (-21 + 11)/(110) = (-10)/(110)`
`v = -(110)/(10) = -11 cm`
Negative sign indicates that the image will be formed on the same side as the object. Therefore, image will be real and inverted.
