**Question:**

Use the mirror equation to deduce that:

(a) an object placed between *f *and 2*f *of a concave mirror produces a real image beyond 2*f.*

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[*Note: *This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

**Solution:**

(a) For a concave mirror, the focal length (*f*) is negative.

∴*f *< 0

When the object is placed on the left side of the mirror, the object distance (*u*) is negative.

∴*u* < 0

For image distance *v*, we can write the lens formula as:

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

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ ...(1)

The object lies between *f* and 2*f*.

$\therefore 2 f

$\frac{1}{2 f}>\frac{1}{u}>\frac{1}{f}$

$-\frac{1}{2 f}<-\frac{1}{u}<-\frac{1}{f}$

$\frac{1}{f}-\frac{1}{2 f}<\frac{1}{f}-\frac{1}{u}<0$ ...(2)

Using equation (1), we get:

$\frac{1}{2 f}<\frac{1}{v}<0$

$\therefore \frac{1}{v}$ is negative, i.e., $v$ is negative.

$\frac{1}{2 f}<\frac{1}{v}$

$2 f>v$

$-v>-2 f$

Therefore, the image lies beyond 2*f*.

(b) For a convex mirror, the focal length (*f*) is positive.

∴ *f *> 0

When the object is placed on the left side of the mirror, the object distance (*u*) is negative.

∴ *u *< 0

For image distance *v*, we have the mirror formula:

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

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

Using equation (2), we can conclude that:

$\frac{1}{v}<0$

$v>0$

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

(c) For a convex mirror, the focal length (*f*) is positive.

∴*f* > 0

When the object is placed on the left side of the mirror, the object distance (*u*) is negative,

∴*u* < 0

For image distance *v*, we have the mirror formula:

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

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

But we have $u<0$

$\therefore \frac{1}{v}>\frac{1}{f}$

$v

Hence, the image formed is diminished and is located between the focus (*f*) and the pole.

(d) For a concave mirror, the focal length (*f*) is negative.

∴*f* < 0

When the object is placed on the left side of the mirror, the object distance (*u*) is negative.

∴*u* < 0

It is placed between the focus (*f*) and the pole.

$\therefore f>u>0$

$\frac{1}{f}<\frac{1}{u}<0$

$\frac{1}{f}-\frac{1}{u}<0$

For image distance *v*, we have the mirror formula:

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

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

$\therefore \frac{1}{y}<0$

$v>0$

The image is formed on the right side of the mirror. Hence, it is a virtual image.

For *u* < 0 and *v* > 0, we can write:

$\frac{1}{u}>\frac{1}{v}$

$v>u$

Magnification, $m=\frac{v}{u}>1$

Hence, the formed image is enlarged.

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