Acids, Bases and Salts

(i) Magnification is positive for a virtual image formed by a lens.

(ii) Magnification is negative for a real image formed by a lens.

Convex and concave lens can be distinguished by touching them. If the curved surface is bulging outwards then it's convex lens and if the curved surface is curved inwards then it's concave lens.

Another way to differentiate between the two lenses is, by bringing some written matter just in front of both the lenses one by one and, look for its image from the other side of the lenses.

(i) If the image of the written matter formed by the lens is virtual, erect and enlarged, then it is a convex lens.

(ii) If the image formed is virtual, erect but diminished, then it is a concave lens.

A concave lens always forms a virtual, erect image on the same side of the object. 

Given,
Image distance, v = -10 cm
Focal length,    f = - 15 cm            [f is -ve for a concave lens]
Object distance, u = ? 

Now, using lens formula, 

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

i.e.,                    $\mathrm{u} = -30 \mathrm{cm}$ 

Thus the object should  be placed at a distance of 30 cm from the lens on the left side. 

Now, 

Magnification, $\mathrm{m}= \frac{\mathrm{v}}{\mathrm{u}} = \frac{-10}{-30} = +\frac{1}{3} = + 0.33$ 

Since, magnification is positive, we can say that the image is erect and virtual.
The size of the  image is reduced to one-third in size than the object after refraction.

Given a convex lens. 

Here,
Focal length, f = +25 cm.
Image distance, v = + 75 cm

By lens formula,
  
                     $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} = \frac{1}{\mathrm{f}}$

$\therefore$                  $$\begin{gathered} \frac{1}{\mathrm{u}} = \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{f}} \\ =\frac{1}{75}-\frac{1}{25} \\ = \frac{1-3}{75} \\ = -\frac{2}{75} \end{gathered}$$ 

i.e.,                  $$\begin{gathered} \mathrm{u} = -\frac{75}{2} \\ = -37.5 \mathrm{cm}. \end{gathered}$$  

The object is at 37.5 cm from the lens and the image is real and inverted.