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.