Question: Q. 6. You are given three lenses $L_{1}, L_{2}$ and $L_{3}$ each of focal length $20 \mathrm{~cm}$. An object is kept at $40 \mathrm{~cm}$ in front of $L_{1}$. The final real image is formed at the focus $I$ of $L_{3}$. Find the separations between $L_{1^{\prime}} L_{2}$ and $L_{3}$.

[O.D. I, II, III 2012]

$$ \begin{aligned} f_{3} & =+20 \mathrm{~cm}, v_{3}=20 \mathrm{~cm} \ \frac{1}{20} & =\frac{1}{20}+\frac{1}{u_{3}} \end{aligned} $$

$\Rightarrow \quad u_{3}=\infty$

$1 / 2$

It shows that $L_{2}$ must render the rays parallel to the common axis. It means that the image $\left(I_{1}\right)$, formed by $L_{1}$, must be at a distance of $20 \mathrm{~cm}$ from $L_{2}$ (at the focus of $L_{2}$ )

Therefore, distance between $L_{1}$ and $L_{2}(=40+20)$ $=60 \mathrm{~cm}$ and distance between $L_{2}$ and $L_{3}$ can have any value.

[CBSE Marking Scheme 2012] 1/22

Detailed Answer :

Given,

$$ f_{1}=f_{2}=f_{3}=20 \mathrm{~cm} $$

For lens, $L_{1}$

$$ \begin{aligned} u & =-40 \mathrm{~cm} \ f & =20 \mathrm{~cm} \ \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \ \frac{1}{v} & =\frac{1}{u}+\frac{1}{f} \ \frac{1}{v} & =-\frac{1}{40}+\frac{1}{20} \ \frac{1}{v} & =\frac{1}{40} \ v & =40 \mathrm{~cm} \end{aligned} $$

( + ve sign shows it is right hand side of lens $L_{1}$ )

Now for $L_{3}$, the final image is at its focus, that means $v_{3}=+20 \mathrm{~cm}$. Hence $u_{3}=\infty$

Now, since image of the object $A B$ formed by convex lens $L_{2}$ is virtual object for $L_{3}$, therefore $v_{2}=\infty$.

Hence for lens $L_{2}, u_{2}=?, f_{2}=20 \mathrm{~cm}$ and $v_{2}=\infty$. Using the lens formula,

$$ \begin{aligned} \frac{1}{v_{2}}-\frac{1}{u_{2}} & =\frac{1}{f_{2}} \ \Rightarrow \quad \frac{1}{\infty}-\frac{1}{u_{2}} & =\frac{1}{20} \ u_{2} & =-20 \mathrm{~cm} \end{aligned} $$

So, the separation between $L$ and $L_{2}$

$$ =40+20=60 \mathrm{~cm} $$

As $v_{2}=\infty$ and $u_{3}=\infty$, therefore the distance between $L_{2}$ and $L_{3}$ does not matter it may take any value because image by $L_{2}$ is formed at infinity.

(AI Q. 7. A convex lens of focal length $20 \mathrm{~cm}$ is placed coaxially with a convex mirror of radius of curvature $20 \mathrm{~cm}$. The two are kept at $15 \mathrm{~cm}$ from each other. A point object lies $60 \mathrm{~cm}$ in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed. [O.D. I, II, III 2014]