Thin Lenses — Question 2

Question 2

A diverging lens gives a reduced virtual image of a real object

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 diverging lens has f=-15 cm and a real object at p=45 cm.

1. Find q and m.

2. Locate the image relative to the lens and near focal point, and find its distance from the object.

Question 2 - Solution

A diverging lens gives a reduced virtual image of a real object

(a) Method and calculation

$$1/q=-1/15-1/45=-4/45\Rightarrow\boxed{q=-11.25\,\mathrm{cm}}.$$$$\boxed{m=-q/p=+0.25}.$$

(b) Interpretation and further calculation

$$|q|=11.25<|f|=15\Rightarrow\boxed{\text{upright virtual image between lens and near focus}}.$$$$\boxed{d_{OI}=45-11.25=33.75\,\mathrm{cm}}.$$

Physical check

The negative image distance places the virtual image on the incident side, closer to the lens than the real object.