A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change in its focal length when it is immersed in water of refractive index 4/3. 

Given, a convex lens.
Refractive index of lens,   = 1.5 
Focal length of lens in air, f_a = 18 
Refractive index of water, ${}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}} = \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

For the lens in air,
               $\frac{1}{{\mathrm{f}}_{\mathrm{a}}} = ({}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}}-1) \left[\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right]$ 

              $\frac{1}{18}= (1.5 - 1) \left[\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right]$ 

               $\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}= \frac{1}{4}$ 

When the lens is immersed in water
               $\frac{1}{{\mathrm{f}}_{\mathrm{w}}} = \left(\frac{{}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}}}{{}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}}}-1\right) \left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)$ 

                    $$\begin{gathered} = \left(\frac{1.5}{4/3}-1\right) \times \frac{1}{4} \\ = \frac{1}{8}\times \frac{1}{4} = \frac{1}{32} \end{gathered}$$ 

Thus,        ${\mathrm{f}}_{\mathrm{w}} = 32 \mathrm{cm}.$  

Hence, focal length changes from 18 to 32.

Suppose I is the real image of an object O. Let d be the distance between them. If the image distance is x, the object distance will be (d – x).
Thus,    u = – (d – x) and v = + x
Sustituting in the lens formula we have
                  
or,                
or,              

For a real image, the value of x must be real, i.e., the roots of the above equation must be real. This is possible if
d² ≥ 4fd
or,    d ≥ 4f
Hence, 4f is the minimum distance between the object and its real image formed by a convex lens.

We have, 

$$\begin{gathered} \mathrm{object} \mathrm{distance}, \mathrm{u} = -15 \mathrm{cm} \\ \mathrm{Focal} \mathrm{length}, \mathrm{f} = -5 \mathrm{cm} \\ \mathrm{Image} \mathrm{distance}, \mathrm{v} = ? \end{gathered}$$ 

Using formula, 

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

$$\begin{gathered} \Rightarrow \mathrm{v} = \frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \\ =\frac{(-15) (-5)}{-15+5} \\ = -7.5 \mathrm{cm} \end{gathered}$$

The image is formed at a distance of 7.5 cm in front of mirror. 
Now,
Magnification, $\mathrm{m} = \frac{\mathrm{I}}{\mathrm{O}} = -\frac{\mathrm{v}}{\mathrm{u}}$

where,
I is the image size and, 

O is the object size. 

       $$\begin{gathered} \frac{\mathrm{I}}{\mathrm{O}}=-\frac{(-7.5)}{(-15)} \\ \mathrm{I} =-0.1 \mathrm{cm} \end{gathered}$$ 

The negative sign indicates that the image is inverted and is diminished in size.

We have,
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm 

We know, using the mirror formula,
                     $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} =\frac{1}{\mathrm{f}}$

$\Rightarrow$                   $$\begin{gathered} \mathrm{v} =\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \\ =\frac{(-30) (+20)}{-30-20} \\ = + 12 \mathrm{cm} \end{gathered}$$ 

The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror. 

Now, size if the object, O = 0.5 cm 
sixe of the image = I
Magnification,  $\mathrm{m} = \frac{\mathrm{I}}{\mathrm{O}} = -\frac{\mathrm{v}}{\mathrm{u}}=-\frac{12}{-30}$
$\Rightarrow$                             $$\begin{gathered} \mathrm{I} = \frac{2}{5}\times \mathrm{O} \\ = \frac{2}{5}\times 0.5 \\ = +0.2 \mathrm{cm} \end{gathered}$$ 

Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size