Question

A diverging lens (f = –11.0 cm) is located 24.0 cm to the left of a converging lens (f = 35.0 cm). A 3.70-cm-tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

Answer 1

Use thin lens equation for the diverging lens:

$$\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f_1},\\\space\\ v_1=\bigg(\frac{1}{-11}-\frac{1}{11}\bigg)^{-1}=-5.5\text{ cm}.$$

The magnification is

$$m_1=v_1/u_1=-5.5/11=-0.5.$$

The height is

$$h_{i1}=0.5\cdot3.7=1.85\text{ cm}.$$

$$\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f_2},\\\space\\ \frac{1}{v_2}+\frac{1}{d+|v_1|}=\frac{1}{f_2},\\\space\\ v_2=\bigg(\frac{1}{35}-\frac{1}{24+5.5}\bigg)^{-1}=-187.7\text{ cm}.$$

Magnification:

$$m_2=-v_2/u_2=187.7/29.5=6.36.$$

Final image height:

$$h_{i2}=m_2h_{i1}=6.36\cdot1.85=11.8\text{ cm}.$$