Optics 3 Question 6
6. What is the position and nature of image formed by lens combination shown in figure? (where, $f _1$ and $f _2$ are focal lengths)
(2019 Main, 12 Jan I)
(a) $\frac{20}{3} cm$ from point $B$ at right, real
(b) $70 cm$ from point $B$ at right, real
(c) $40 cm$ from point $B$ at right, real
(d) $70 cm$ from point $B$ at left, virtual
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Answer:
Correct Answer: 6. (b)
Solution:
- For lens $A$, object distance, $u=-20 cm$
Focal length, $f=+5 cm$
From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
We have, $\quad \frac{1}{v}=\frac{1}{5}-\frac{1}{20}$
$$ \frac{1}{v}=\frac{20-5}{20 \times 5}=\frac{15}{100} $$
$$ v=\frac{20}{3} cm $$
For lens $B$, image of $A$ is object for $B$.
$$ \therefore \quad u=\frac{20}{3}-2=+\frac{14}{3} cm $$
$$ f=-5 cm $$
Now, from lens formula, we have
$$ \begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \frac{1}{v} & =\frac{1}{-5}+\frac{3}{14} \\ \frac{1}{v} & =\frac{15-14}{5 \times 14} \\ v & =70 cm \end{aligned} $$
Hence, image is on right of lens $B$ and is real in nature.
