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# The image formed by a spherical mirror is real, inverted and is of magnification â€“2. If the image is at a distance of 30 cm from the mirror, where is the object placed? Find the focal length of the mirror. **List two characteristics of the image formed if the object is moved 10 cm towards the mirror.**

As the image formed by the spherical mirror is real, inverted, and has magnification $-2$, it means the spherical mirror is concave in nature. Because only a concave mirror forms a real and inverted image.

**
**

**Given:**

**Concave Mirror **

Magnification, $m=-2$

Image distance, $v=-30cm$ (image is taken negative, as it forms on the left side of the mirror)

**To find: **Position of the object, $u$, and focal length of the mirror, $f$.

**
**

**Solution: **

From the magnification formula, we know that-

$m=\frac {-v}{u}$

Putting the give values, we get-

$-2=\frac {-(-30)}{u}$

$u=\frac {30}{-2}$

$u=-15cm$

Thus, the position of the object is **15 cm** in front of the mirror.

Now,

From the mirror formula we know that-

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

Substituting the value of $v$ and $u$ we get-

$\frac {1}{f}=\frac {1}{(-30)}+\frac {1}{(-15)}$

$\frac {1}{f}=-\frac {1}{30}-\frac {1}{15}$

$\frac {1}{f}=\frac {-1-2}{30}$

$\frac {1}{f}=\frac {-3}{30}$

$\frac {1}{f}=-\frac {1}{10}$

$f=-10cm$

Thus, the focal length of the mirror is **10 c**m.

The object distance is **15cm** and the focal length is **10cm**, which means the object placed between focus $(F)$ and centre of curvature $(C\ or\ 2F)$.

Further, according to the question if the object is moved **10 cm** towards the mirror, then the object distance will become-

$u=15-10=5cm$

This means the object is now placed within the focus $(F)$, or between pole $(P)$ and focus $(F)$.

We know that when the object is placed within the focus $(F)$, then the image formed by the concave mirror is virtual, erect and magnified.

Therefore, the two characteristics of the image formed if the object is moved 10 cm towards the mirror (between pole, $P$ and focus, $F$) are as follows:

**1.** Image will be virtual.

**2.** Image will be erect.