Question

Light from a burning match propagates from left to right, first through a thin lens of focal length 5.1 cm, and then through another thin lens, with a 9.7-cm focal length. The lenses are fixed 32.8 cm apart. A real image of the flame is formed by the second lens at a distance of 22.6 cm from the lens.
(a) How far from the second lens, in centimeters, is its optical object located?
(b) How far is the burning match from the first lens, in centimeters?

Answer 1

Given,

Focal length of the thin lens $(f_1)=5.1cm$

Focal length of the second thin lens $(f_2)=9.7cm$

Distance between the focal length $(x)=32.8cm$

Real image of the flame is formed by the second lens at a distance of $(d_{i2})=22.6cm$

From the lens equation,

$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$

Substituting for the second lens

$\frac{1}{d_{o2}}+\frac{1}{d_{i2}}=\frac{1}{f_2}$

$\Rightarrow \frac{1}{d_{o2}}+\frac{1}{22.6}=\frac{1}{9.7}$

$\Rightarrow \frac{1}{d_{o2}}=\frac{1}{9.7}-\frac{1}{22.6}$

$d_{o2}=16.67cm$

Hence the optical object is located at 16.67cm

b) As the lens are separated by 32.8cm hence, the image distance for the first lens $=32.8cm-16.67 cm =16.13cm$

$\frac{1}{d_o}+\frac{1}{d_{i1}}=\frac{1}{f_1}$

$\frac{1}{d_o}=\frac{1}{f}-\frac{1}{d_i}$

$\frac{1}{d_o}=\frac{1}{5.1}-\frac{1}{16.13}$

$d_o=7.46cm$

Hence the burning match from the first lens is 7.46cm