The position of final image formed by the given lens combination from the third lens will be at a distance of `(f_(1)=+10 cm, f_(2)=-10cm, f_(3)=+30cm)`

Options

A. 15 cm
B. infinity
C. 45 cm
D. 30 cm

Answer

D

Explanation

For first lens, `u_(1)=-30cm, f_(1)=10cm`
`(1)/(f)=(1)/(v)-(1)/(u)`
or `(1)/(v)=(1)/(f)+(1)/(u)`
or `(1)/(v)=(1)/(10)-(1)/(30)=(1)/(15)`
Therefore, image formed by convex lens `(L_(1))` is at point `I_(1)` and acts as virtual object for concave lons `(L_(2))`.
`f_(1)=10cm f_(2)=10cm f_(3)=30cm`

The image `I_(1)` is formed at focus of concave lens (as shown) and so emergent rays will be parallel to the principle axis. For lens `L_(2),u_(2)=15-5=10cm, f_(2)=-10cm`
These parallel rays are incident on the third convex lens `(L_(3))` and will be brought to convergence at the focus of the lens `L_(3)`.
Hence, distance of final image from third lens `L_(3)`,
`v_(2)=f_(3)=30cm`