(a) We know for a concave mirror f < 0 [negative] and u < 0 [negative]
2f < u < f
∴

1

2f

>

1

u

>

1

f

or −

1

2f

<

−1

u

<

−1

f

or

1

f

−

1

2f

<

1

f

−

1

u

<

1

f

−

1

f

or

1

2f

<

1

v

< 0 [Since

1

f

−

1

u

=

1

v

]
Which implies that v < 0 to form image on the left.
Also 2f > v {Since 2f and v are –ve}
|2f| < |v|
So, the real image is formed beyond 2f.
(b) For a convex mirror, f > 0, always positive and object distance u < 0, always negative.
Now mirror formula,

1

f

=

1

v

+

1

u

or

1

v

=

1

f

−

1

u

This implies that

1

v

> 0 , or v > 0.
So, whatever be the value of u, a convex mirror always forms a virtual image.
(c) In convex mirror focal length is positive hence f > 0 and for an object distance from mirror with negative sign (u < 0)
So,

1

v

+

1

u

=

1

f

or

1

v

=

1

f

−

1

u

The results

1

v

>

1

f

or v < f (both positive) hence the image is located between pole and focus of the mirror.
Also magnification m = -

v

u

= -

+v

−u

m < [1] (positive)
So, the image is virtual and diminished.
(d) In concave mirror, f < 0 for object placed between focus and pole of concave mirror
f < u < 0 (both negative)

1

f

>

1

u

Now mirror formula,

1

v

=

1

f

−

1

u

1

v

> 0 or v > 0 (positive)
hence the image is virtual.
Also magnification m = -

v

u

, here

1

v

<

1

|u|

or v > |u|
So, m > 1 hence the image is enlarged.