Given,
Power of the lens, P = -2.0 D

Therefore, 
Focal length, $\mathrm{f} = \frac{1}{\mathrm{P}} = \frac{1}{-2.0 \mathrm{D}} = -0.5 \mathrm{m}.$

Since, focal length is negative, the lens is concave.

Focal length, f = - 15 cm    [f is - ve for a concave lens]
Image distance, v = - 10 cm [Concave lens forms virtual image on same side as the object, so v is - ve]
As,                                    

         
              Object distance, 

Drawing the ray diagram: Using a scale of 1: 5, we get v = - 2 cm, f = - 3 cm. We draw the ray diagram as follows:
(i) Draw the principal axis (a horizontal line).
(ii) Draw a convex lens, keeping principal centre (C) on the principal axis.
(iii) Mark points F and B on the left side of lens at a distance of 3 cm and 2 cm respectively.
(iv) Join any point D (nearly at the top of lens) and F by a dotted line.
(v) Draw a line AD, parallel to principal axis.

(vi) Draw a line A'B', perpendicular to principal axis from B'.
(vii) Draw a line CA', backwards, so that it meets the line from D parallel to principal axis at A.
(viii) Draw a line AB, perpendicular (downwards) from A to meet the principal axis at B.
(ix) The AB is position of object. Measure distance BC. It will be found to be equal to 6 cm.
Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens.

Magnification is the ratio of the size of the image (h') to the size of the object(h). 

As,
Magnification, $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}}$

For a plane mirror,  m = + 1 (given).
So, 
                  h' = h and  v = -u 

Magnification is equal to one indicates that the size of image is same as that of object.
Positive sign of m indicates that a virtual image is formed behind the mirror.

We are given a convex mirror.

Here, we have 

Object size, h = + 5 cm 
Object distance, u = -20 cm
Radius of curvature, R = + 3.0 cm [R is +ve for a convex mirror]
$\therefore$ Focal length ,  $\mathrm{f} = \frac{\mathrm{R}}{2} = +15 \mathrm{cm}$ 

From mirror formula,

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

we have, 

                      $$\begin{gathered} \frac{1}{v}= \frac{1}{+15}-\frac{1}{-20} \\ = \frac{4+3}{60} \\ = \frac{7}{60} \end{gathered}$$ 

Image distance, $\mathrm{v} = \frac{60}{7}\simeq 8.6 \mathrm{cm}.$ 

Magnification, $\mathrm{m} = -\frac{\mathrm{v}}{\mathrm{u}}= \frac{\mathrm{h}\text{'}}{\mathrm{h}}$ 
Therefore, 

 $$\begin{gathered} \mathrm{Image} \mathrm{size}, \mathrm{h}\text{'} = -\frac{\mathrm{vh}}{\mathrm{u}} \\ = -\frac{8.6 \times 5}{-20} \\ = 2.15 \simeq 2.2 \mathrm{cm}. \end{gathered}$$ 

A virtual and erect image of height 2.2 cm is formed behind the mirror (because v is positive) at a distance of 8.6 cm from the mirror. 

We are given a convex mirror. 
Here, 

Object distance, u = -10 cm 
Focal length, f = + 15 cm [f is +ve for a convex mirror]
Image distance,  v = ? 

Using the mirror formula, 

                     $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}$ 
we have, 

$\therefore$                  $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{+15}-\frac{1}{-10} \\ = \frac{2+3}{30} \\ =\frac{1}{6} \end{gathered}$$ 

Thus, image distance,  v = + 6 cm.
 
As image distance is +ve, so a virtual, erect image is formed at a distance 6 cm behind the mirror.