Question

Two thin biconvex lenses L1 and L2 having focal lengths 10 cm and 15 cm respectively are placed coaxialy. What distance (d) should be maintained between them so that a parallel beam of light parallel to their axis, incident on the lense L1 emerges from L2 undeaviated? If the distance between them is increased by d, at what distance from L1 on the axis, does the emergent beam converges?

Answer 1

Dear Student,

The minimum distance between lens L1 and L2 will be sum of their focal lengths. i.e 10+15 =25 cm.
As for a parallel beam of light incident on lens L1, the rays will focus on its focal length, 10 cm from the lens. Now, for lens L2, the point ray source kept at its focus will be refracted parallel along its axis, so it should converge 10 cm from the lens L1 , which is 15 cm before lens L2. This makes the distance between both of them as 25 cm.

If the distance between L1 and L2 lens is increased by d.
Then, object distance for lens 2 will be, $u_2=-\left(f_2+d\right)$
using lens formula , Image distance for L2 will be
$$\begin{gathered} \frac{1}{v_2}=\frac{1}{f_2}+\frac{1}{u_2} \\ \Rightarrow \frac{1}{v_2}=\frac{1}{f_2}+\frac{1}{-\left(f_2+d\right)} \\ \Rightarrow \frac{1}{v_2}=\frac{d}{f_2\left(f_2+d\right)} \\ \Rightarrow v_2=\frac{f_2\left(f_2+d\right)}{d} \\ \Rightarrow v_2=\frac{15\left(15+d\right)}{d} \end{gathered}$$
Distance between L1 and image of L2 is 10 + 15 +d + v₂
$$\begin{gathered} \Rightarrow 10+15+d+v_2=25+d+\frac{15\left(15+d\right)}{d} \\ \Rightarrow \frac{d^2+40d+225}{d} \end{gathered}$$

Regards