ray-optics-and-optical-instruments Question 36
Question: Q. 5. In the following diagram, an object ’ $O$ ’ is placed $15 \mathrm{~cm}$ in front of a convex lens $L_{1}$ of focal length 20 $\mathrm{cm}$ and the final image is formed at ’ $I$ ’ at a distance of $80 \mathrm{~cm}$ from the second lens $L_{2}$. Find the focal length of the lens $L_{2}$.
[Foreign 2016]
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Solution:
Ans. For lens $\mathrm{L}_{1}$
Using lens formula
$$ \begin{aligned} u & =-15 \mathrm{~cm} \ f & =+20 \mathrm{~cm} \end{aligned} $$
$$ \begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \ \frac{1}{v} & =\frac{1}{u}+\frac{1}{f} \ \frac{1}{v} & =-\frac{1}{15}+\frac{1}{20}=-\frac{1}{60} \ v & =-60 \mathrm{~cm} \end{aligned} $$
This image will act as object for lens $L_{2}$ Hence for lens $L_{2}$
$$ \begin{align*} & u=-20-60=-80 \mathrm{~cm} \ & v=+80 \mathrm{~cm} \ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \ & \frac{1}{80}+\frac{1}{80}=\frac{1}{f} \ & \frac{1}{f}=\frac{1}{40} \ & f=40 \text { cm } \tag{1}\ & \text { [CBSE Marking Scheme 2016] } \end{align*} $$
$$ \text { or } \quad f=40 \mathrm{~cm} $$
