An object is placed `12 cm` to the left of a converging lens of focal length `8 cm`. another converging lens of `6cm` focal length is placed at a distance of `30 cm` to the right of the first lens. The second lens will produce

Options

A. a virtual enlarged image
B. no image
C. a real inverted image
D. a real enlarged image.

Answer

C

Explanation

AB is the object and A.B. is the image formed by the first lens `L_(1)`.
Image distance (v) is calculated from the relation,
`(1)/(v)-(1)/(u)=(1)/(f)` (Lens formula)
Here, u=+12cm, f=-8cm.
[Distance measured along the direction of the incident ray is taken as negative].
`therefore(1)/(v)-(1)/(12)=-(1)/(8)impliesv=-24cm`.
So the first lens `L_(1)` will form a real image at a distance of 24 cm to the right of it. this image will act as the object for the second lens `L_(2)`.
Now, the object distance for the second lens
`L_(2)=30-24=6cm`.
Since the focal length of the `2^(nd)` lens is 6cm, the object is now placed at the focus of the `2^(nd)` lens. hence the image will be formed at infinity which will be real and inverted.