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List the sign conventions for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three times magnified real image of an object placed 16 cm in front of it.
Sign conventions for spherical mirrors:
1. All distances are measured from the pole of the mirror as origin.
2. Objects are always placed to the left of the mirror.
3. Distances measured in the same direction as that of incident light (along positive X - axis) are taken as positive.
4. Distances measured against the direction of incident light (along negative X-axis) are taken as negative.
5. Heights measured upward and perpendicular to the principal axis (along positive Y-axis) are taken as positive.
6. Heights measured downward and perpendicular to the principal axis (along negative Y-axis) are taken as negative.
Given:
Magnificatio, $m$ = $-$3 (Real image is always inverted)
Distance of the object from the mirror, $u$ = $-$16 cm
To find: Focal length of the mirror, $(f)$.
Solution:
From the magnification formula we know that-
$m=-\frac {v}{u}$
Substituting the given values we get-
$-3=-\frac {v}{(-16)}$
$v=-48cm$
Thus, the distance of the image $v$ from the spherical mirror is 48 cm, and the negative sign implies that it forms in front of the mirror (on the left).
Now, from the mirror formula, we know that-
$\frac {1}{f}=\frac {1}{v}+\frac {1}{u}$
Substituting the given values we get-
$\frac {1}{f}=\frac {1}{(-48)}+\frac {1}{(-16)}$
$\frac {1}{f}=-\frac {1}{48}-\frac {1}{16}$
$\frac {1}{f}=-\frac {1}{48}-\frac {1}{16}$
$\frac {1}{f}=\frac {-1-3}{48}$
$\frac {1}{f}=\frac {-4}{48}$
$\frac {1}{f}=\frac {-1}{12}$
$f=-12cm$
Thus, the focal length of the spherical mirror is 12 cm, and the negative sign implies that it is in front of the mirror (on the left), which means the mirror is concave in nature.
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