Question

A thin converging lens L₁ forms a real image of an object located far away from the lens as shown in figure. The image is located at a distance 4l and has height h. A diverging lens of focal length l is placed at a distance 2l from lens L₁ at A. Another converging lens of focal length 2l is placed at a distance 3l from lens L₁ at B. The height of final image thus formed is –

• A. h
• B. 4h
• C.
• D. 2h

Answer 1

Answer: (D)

2h

Answer 2

For L₂ :

$\frac { 1 } { v _ { 2 } } - \frac { 1 } { - 2 \ell }$ = $\frac { 1 } { - \ell }$

v₂ = –2l

For L₃ :

$\frac { 1 } { v _ { 3 } } - \frac { 1 } { - 3 \ell }$= $\frac { 1 } { 2 \ell }$

Ž v₃ = 6l

m₂ = =$\frac { - 2 \ell } { 2 \ell }$= –1

m₃ = = – 2

Total magnification = m₂ m₃ = 2

\ Object of height h get magnified two times.