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A concave mirror is used for image formation for different positions of an object. What inferences can be drawn about the following when an object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm?(a) Position of the image(b) Size of the image(c) Nature of the imageDraw a labeled ray diagram to justify your inferences.
Given:
Object distance, $u$ = $-$10 cm (object distance will be negative, as it is placed on the left side of the mirror)
Focal length, $f$ = $-$15 cm (focal length of the concave mirror is taken negative)
To find: (a) Position of the image.
(b) Size of the image.
(c) Nature of the image.
Solution:
From the mirror formula we know that-
$\frac {1}{v}+\frac {1}{u}=\frac {1}{f}$
Substituting the given values, we get-
$\frac {1}{v}+\frac {1}{(-10)}=\frac {1}{(-15)}$
$\frac {1}{v}-\frac {1}{10}=-\frac {1}{15}$ $\frac {1}{v}=\frac {1}{10}-\frac {1}{15}$
$\frac {1}{v}=\frac {3-2}{30}$
$\frac {1}{v}=\frac {1}{30}$
$v=+30cm$
Thus, the position of the image, $v$ is 30 cm from the mirror. The positive sign implies that the image is formed behind the mirror (on the right side).
As $v$ is positive the nature of the image will be virtual and erect.
Now,
From the magnification formuala we know that-
$m=-\frac {v}{u}$
$m=-\frac {30}{-10}$
$m=+3$
Thus, the magnification is 3, which means that the image will be highly magnified in size.
Hence,
(a) Position of the image - 30cm behind the mirror.
(b) Size of the image - highly magnified in size.
(c) Nature of the image - virtual and erect.
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