# Use the mirror equation to deduce that:

Question:

Use the mirror equation to deduce that:

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

(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.

< 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 2f.

$\therefore 2 f$\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$Therefore, the image lies beyond 2f. (b) For a convex mirror, the focal length (f) is positive. ∴ > 0 When the object is placed on the left side of the mirror, the object distance (u) is negative. ∴ < 0 For image distance v, we have the mirror formula:$\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}<0v>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.