Thin Lenses — Question 1

Question 1

A projection lens fixes screen position and inverted image size

Use the thin-lens relations $1/p+1/q=1/f$ and $m=-q/p$ in one surrounding medium. Distances p and q are positive for real objects and real images; f is positive for a converging lens. A lens of f=12 cm views a 5-mm object at p=18 cm.

1. Find the screen position for a sharp image and its magnification.

2. Find signed image height and total object-screen separation.

Question 1 - Solution

A projection lens fixes screen position and inverted image size

(a) Method and calculation

$$q=fp/(p-f)=12(18)/6=\boxed{36\,\mathrm{cm}}.$$$$\boxed{m=-36/18=-2}.$$

(b) Interpretation and further calculation

$$\boxed{h_I=-2(5)=-10\,\mathrm{mm}}.$$$$\boxed{L=p+q=54\,\mathrm{cm}}.$$

Physical check

For transmission through a lens, the real object and real image lie on opposite sides, so their physical separation is p+q.