In a concave mirror, an object is placed at a distance $d_{1}$ from the focus and a real image is formed at a distance $d_{2}$ from the focus. Find the focal length of the mirror?

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Solution

In a concave mirror, object distance and image distance is negative. So

$u = - (f + d_{1})$
$v = - (f + d_{2})$

$F = - f$

Using mirror’s relation

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{- (f + d_{1})} + \frac{1}{- (f + d_{2})} = - \frac{1}{f}$

$\frac{1}{(f + d_{1})} + \frac{1}{(f + d_{2})} = \frac{1}{f}$

$\frac{f + d_{1} + f + d_{2}}{(f + d_{1}) (f + d_{2})} = - \frac{1}{f}$

$2 f^{2} + f d_{1} + f d_{2} = f^{2} + f d_{2} + f d_{1} + d_{1} d_{2}$

$f^{2} = d_{1} d_{2}$

$f = \sqrt{d_{1} d_{2}}$

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In a concave mirror, an object is placed at a distance d1 from the focus and a real image is formed at a distance d2 from the focus. Find the focal length of the mirror?

In a concave mirror, object distance and image distance is negative. So u=−(f+d1)v=−(f+d2) F=−f Using mirror’s relation 1v+1u=1f 1−(f+d1)+1−(f+d2)=−1f 1(f+d1)+1(f+d2)=1f f+d1+f+d2(f+d1)(f+d2)=−1f 2f2+fd1+fd2=f2+fd2+fd1+d1d2 f2=d1d2 f=√d1d2

In a concave mirror, an object is placed at a distance $d_{1}$ from the focus and a real image is formed at a distance $d_{2}$ from the focus. Find the focal length of the mirror?