(a) Ray diagram showing the formation of image of an object placed between infinity and the optical centre $(C)$ of a concave lens.
When an object is placed between infinity and the optical centre $(C)$ of a concave lens, then the image formed is virtual, erect, highly diminished, and in front of the lens.
(b) Given:
Focal length, $f$ = $-$15 cm (focal length of a concave lens is always taken negative)
Image distance, $v$ = $-$10 cm (as the image is formed on the left side of the lens, it is taken negative)
To find: (i) The distance of the object from the lens, $u$.
(ii) The magnification for the image formed, $m$.
(iii) The nature of the image formed.
Solution:
(i) From the lens formula we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values we get-
$\frac {1}{(-10)}-\frac {1}{u}=\frac {1}{(-15)}$
$-\frac {1}{10}-\frac {1}{u}=-frac {1}{15}$
$\frac {1}{u}=\frac {1}{15}-\frac {1}{10}$
$\frac {1}{u}=\frac {2-3}{30}$
$\frac {1}{u}=-\frac {1}{30}$
$u=-30cm$
Thus, the object, $u$ is at a distance of 30cm from the concave lens (on the left side).
(ii) From the magnification formula we know that-
$m=\frac {v}{u}$
Substituting the value of $v$ and $u$ we get-
$m=\frac {-10}{-30}$
$m=\frac {1}{3}$
$m=+0.33$
Thus, the magnification, $m$ for the image formed is +0.33.
(iii) The nature of the image formed is virtual, erect, and diminished. Since the value of magnification is less than 1 (it is 0.33), therefore the image is smaller than the object (or diminished). The plus sign $(+)$ for the magnification implies that the image is virtual and erect.
Also, we know that a concave lens always forms virtual, erect, and diminished images.