Answer to Question #132942 in Optics for stephanie

Question

suppose that two converging lens of focal length 10cm and 20cm are separated by 20.0cm.An object is placed 15.0cm to the left of lens 1. What is the position o the final image and magnification?

Answer 1

$f_1 = 10 \;cm$

$f_2 = 20 \;cm$

$d = 20 \;cm$

$u = -15 \;cm$

Image formed by the first lens:

$\frac{1}{f_1} = \frac{1}{v} – \frac{1}{u}$

$\frac{1}{10} = \frac{1}{v} – \frac{1}{-15}$

$v = 30 \;cm$

Object distance for the second lens:

$u’ = v – d$

$u’ = 30 – 20 = 10\; cm$

Final image:

$\frac{1}{f_2} = \frac{1}{v’} - \frac{1}{u’}$

$\frac{1}{-20} = \frac{1}{v’} - \frac{1}{10}$

$v’ = 20 \;cm$

Total magnification:

$m = m_1 \times m_2$

$m = \frac{v}{u} \times \frac{v’}{u’}$

$m = \frac{30}{15} \times \frac{20}{10}$

$m = 4$

Answer: 20 cm and 4.