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Consider a concave mirror of $10 \mathrm{~cm}$ focal length illuminated by an object kept at a distance of $25 \mathrm{~cm}$. The distance at which the image is formed and its magnification respectively are
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1837 Upvotes
Verified Answer
The correct answer is:
$-16.7 \mathrm{~cm}$ and -0.67
Using mirror equation, we have,
$$
\begin{aligned}
\frac{1}{v}+\frac{1}{u}= & \frac{1}{f} \\
\frac{1}{v}= & \frac{1}{f}-\frac{1}{u} \\
\Rightarrow \quad \frac{1}{v}= & \frac{1}{(-10)}-\frac{1}{(-25)} \\
& {[\therefore \text { Given } f=-10 \mathrm{~cm}, u=-25 \mathrm{~cm}] } \\
\Rightarrow \quad v= & -16.7 \mathrm{~cm}
\end{aligned}
$$
or
and magnification,
$$
m=-\left(\frac{v}{u}\right)=\frac{-(-16.7)}{-25}=-0.67
$$
$$
\begin{aligned}
\frac{1}{v}+\frac{1}{u}= & \frac{1}{f} \\
\frac{1}{v}= & \frac{1}{f}-\frac{1}{u} \\
\Rightarrow \quad \frac{1}{v}= & \frac{1}{(-10)}-\frac{1}{(-25)} \\
& {[\therefore \text { Given } f=-10 \mathrm{~cm}, u=-25 \mathrm{~cm}] } \\
\Rightarrow \quad v= & -16.7 \mathrm{~cm}
\end{aligned}
$$
or
and magnification,
$$
m=-\left(\frac{v}{u}\right)=\frac{-(-16.7)}{-25}=-0.67
$$
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