Use the mirror equation to deduce that: (a) an object placed between f and 2f of 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.]

(a) For a concave mirror, the focal length (f) is negative. Therefore ,f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore  u < 0 For image distance v, we can write the lens formula as:

1/v – 1/u = 1/f

1/v = 1/f – 1/u——————- Eq-1

The object lies between f and 2f.

However , 2f <u <f          (because  u and f are negative)

1/2f > 1/u > 1/f

-1/2f < -1/u <-1/f

1/f -1/2f <1/f  -1/u<0       —————-Eq-2

Using equation (1), we get:

1/2f < 1/v < 0

Therefore, 1/v  is negative, i.e., v is negative.

1/2f < 1/v => 2F > v

=> -v > -2f ,therefore, the image lies beyond 2f.

Ans (b).

For a convex mirror, the focal length (f) is positive. Therefore f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore,  u < 0 For image distance v, we have the mirror formula:

1/v +1/u = 1/f

1/v = 1/f -1/u

Using equation (2), we can conclude that:

1/v<o

0r 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.

Ans (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, therefore, u < 0 For image distance v, we have the mirror formula:

1/v + 1/u = 1/f

=> 1/v = 1/f -1/u

But we have u < 0

Therefore , 1/v > 1/f

or v <f

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

Ans (d).

For a concave mirror, the focal length (f) is negative. Therefore, f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 It is placed between the focus (f) and the pole.

Therefore ,f > u> 0

1/f < 1/u < 0

1/f -1/u < 0

For image distance v, we have the mirror formula:

1/v + 1/u = 1/f

1/v = 1/f -1/u

Therefore ,1/v < 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: 1/u > 1/v => v > u

Magnification, m = v/u > 1 Hence, the formed image is enlarged.