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**(a) **Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens.**(b) **A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate**i. **The distance of the object from the lens.**ii. **The magnification for the image formed**iii. **The nature of the image formed.

**(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.