Question

Two convex lenses, each of focal length 1.0 cm, are placed at a separation of 15 cm with their principal axes coinciding.

(a) Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system.

(b) Find the location of the virtual image formed by the lens system of an object placed far away.

(c) Find the focal length of the equivalent lens. (Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it.)

Answer 1

(a) The beam will diverge after coming out of the convex lens system because, the image formed by the first lens lies within the focal length of the second lens.

(b) For 1st Convex lens,

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

$=\frac{1}{10}+\frac{1}{(-\infty )}=\frac{1}{10}$ [since, $\mu =-\infty$]

or, V = 10 cm

For 2nd convex lens,

$\frac{1}{v_1}=\frac{1}{f}+\frac{1}{u}$

or, $\frac{1}{v_1}=\frac{1}{10}+\frac{1}{-(15-10)}$

$=\frac{1}{10}+\frac{1}{(-5)}$

$=\frac{1}{10}$

or $v_1=-10\text{cm}$

So, the virtual image will be at 5 cm from 1st convex lens.

(c) If F be the focal length of equivalent lens.

Then, $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}$

$=\frac{1}{10}+\frac{1}{10}-\frac{15}{100}$

$=\frac{1}{5}-\frac{3}{20}$

$=\frac{1}{20}$

or $F=20\text{cm}$