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A biconvex lens of focal length $15 \mathrm{~cm}$ is in front of a plane mirror. The distance between the lens and the mirror is 10 $\mathrm{cm}$. A small object is kept at a distance of $30 \mathrm{~cm}$ from the lens. The final image is
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Verified Answer
The correct answer is:
real and at a distance of $16 \mathrm{~cm}$ from the mirror
real and at a distance of $16 \mathrm{~cm}$ from the mirror
Object is placed at distance $2 f$ from the lens. So first image $I_1$ will be formed at distance $2 f$ on other side. This image $I_1$ will behave like a virtual object for mirror. The second image $I_2$ will be formed at distance $20 \mathrm{~cm}$ in front of the mirror, or at distance $10 \mathrm{~cm}$ to the left hand side of the lens.
Now applying lens formula
$$
\begin{array}{rlr}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15} \\
\text { or } & v=6 \mathrm{~cm}
\end{array}
$$
Therefore, the final image is at distance $16 \mathrm{~cm}$ from the mirror. But, this image will be real.
This is because ray of light is travelling from right to left.
$\therefore$ The correct option is (b).
Now applying lens formula
$$
\begin{array}{rlr}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15} \\
\text { or } & v=6 \mathrm{~cm}
\end{array}
$$
Therefore, the final image is at distance $16 \mathrm{~cm}$ from the mirror. But, this image will be real.
This is because ray of light is travelling from right to left.
$\therefore$ The correct option is (b).
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