Thin Lenses — Question 8

Question 8

A second lens collimates a real intermediate image

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. Lens 1 has f1=10 cm and views an object at p1=15 cm. Lens 2, with f2=20 cm, is 50 cm to the right of lens 1. The original object height is 2 mm.

1. Find the intermediate image and show that the final beam is collimated.

2. Find the intermediate signed image height and outgoing beam angle, using $\alpha_{\mathrm{out}}=-h_1/f_2$.

Question 8 - Solution

A second lens collimates a real intermediate image

(a) Method and calculation

$$q_1=10(15)/(15-10)=\boxed{30\,\mathrm{cm}},\quad p_2=50-30=20\,\mathrm{cm}=f_2.$$$$1/q_2=1/f_2-1/p_2=0\Rightarrow\boxed{q_2\to\infty}.$$

(b) Interpretation and further calculation

$$m_1=-30/15=-2\Rightarrow\boxed{h_1=-4\,\mathrm{mm}}.$$$$\boxed{\alpha_{\mathrm{out}}=-(-4)/(200)=+0.020\,\mathrm{rad}}.$$

Physical check

An image at infinity is described by outgoing direction rather than a finite final image height.