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In a concave mirror, an object is placed at a distance $d_1$ from the focus and the image is formed at a distance $d_2$ from the focus. Then the focal length of the mirror is
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Verified Answer
The correct answer is:
$\sqrt{d_1 d_2}$
$\mathrm{U} \operatorname{sing} \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Given $u=f+d_1 \quad v=f+d_2$
$f=\frac{u v}{u+v}=\frac{\left(f+d_1\right)\left(f+d_2\right)}{\left(f+d_1\right)+\left(f+d_2\right)}$
On solving $f=\sqrt{d_1 d_2}$
Given $u=f+d_1 \quad v=f+d_2$
$f=\frac{u v}{u+v}=\frac{\left(f+d_1\right)\left(f+d_2\right)}{\left(f+d_1\right)+\left(f+d_2\right)}$
On solving $f=\sqrt{d_1 d_2}$
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