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]
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Solution:
Ans.
For the convex lens, $u=-60 \mathrm{~cm}, f=+20 \mathrm{~cm}$
$$ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \text { gives } v=+30 \mathrm{~cm} 1 / 2 $$
For the convex mirror,
$$ \begin{gathered} u=+(30-15) \mathrm{cm}=15 \mathrm{~cm}, f=+\frac{20}{2} \mathrm{~cm}=10 \mathrm{~cm}^{1 / 2} \ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \text { gives } v=+30 \mathrm{~cm} \end{gathered} $$
The final image is formed at the distance of $30 \mathrm{~cm}$ from the convex mirror (or $45 \mathrm{~cm}$ from the convex lens) to the right of the convex mirror.
AT Q. 8. A ray $P Q$ is incident normally on the face $A B$ of a triangular prism of refracting angle of $60^{\circ}$, made of a transparent material of refractive index $\frac{2}{\sqrt{3}}$, as shown in the figure. Trace the path of the ray as it passes through the prism. Also calculate the angle of emergence and angle of deviation.
[Delhi Comptt. I, II, III 2014]
Ans. If $i_{c}$ is the critical angle for the prism/material,
Angle of incidence at face $A C$ of the prism $=60^{\circ} \frac{1}{2}$ Hence, refracted ray grazes the surface $A C$.
$\Rightarrow$ Angle of emergence $=90^{\circ}$
$1 / 2$
$\Rightarrow$ Angle of deviation $=30^{\circ}$
[CBSE Marking Scheme 2014] 11/2