Question

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.]

Answer 1

a)

For a concave mirror the focal length f<0 and object distance u<0 are negative when the object is placed left side of the mirror.

The image distance can be calculated using the mirror formula.

1 f = 1 v − 1 u 1 v = 1 f − 1 u (1)

As it is given that object lies between f and 2f then,

1 2f < 1 u < 1 f − 1 2f <− 1 u <− 1 f 1 f − 1 2f < 1 f − 1 u < 1 f − 1 f 1 f − 1 2f < 1 f − 1 u <0

Substituting the value from equation (1), we get

1 2f < 1 v <0(2)

This shows that v is negative.

From equation (2),

2f>v −v>−2f

Hence, from the above expression we can say that the image formed will lie beyond 2f.

b)

For a convex mirror focal length f>0 is positive and the object distance u<0 is negative as it lies to the left of the mirror.

The image distance can be calculated using the mirror formula.

1 f = 1 v − 1 u 1 v = 1 f − 1 u

From equation (2), we can conclude that,

1 v <0 v>0

The above expression states that the image is formed behind the mirror.

Thus, the convex mirror produces a virtual image independent of location of the object.

c)

For a convex mirror focal length f>0 is positive and the object distance u<0 is negative as it lies to the left of the mirror.

The image distance can be calculated using the mirror formula.

1 f = 1 v − 1 u 1 v = 1 f − 1 u

As we have u<0 then,

1 v > 1 f v<f

Thus, the image formed will be diminished and will be located between f and the pole.

d)

For a concave mirror the focal length f<0 and object distance u<0 are negative when the object is placed left side of the mirror.

The image distance can be calculated using the mirror formula.

1 f = 1 v − 1 u 1 v = 1 f − 1 u (3)

The object is placed between f and the pole hence,

f>u>0 1 f < 1 u <0 1 f − 1 u <0

Now, from equation (3), we get

1 f = 1 v − 1 u 1 v = 1 f − 1 u 1 v <0 v>0

Thus, from the above expression we can say that the image will be formed on the right side of the mirror and will be a virtual image.

When u<0 and v>0,

1 u > 1 v v>u (4)

Let m be the magnification then,

m= v u

From equation (3),

m>1

As magnification is greater than 1 therefore, the image formed will be enlarged.