i.) Derive the lens formula, 1/ f = 1/v – 1/u for a concave lens, using the necessary ray diagram.
ii.) How does the angle of minimum deviation of a glass prism of refractive index 1.15 change, if it is immersed in a liquid of refractive index 1.3?

Let, AB be the object held perpendicular to the principal axis at a distance beyond the focal length of the lens. And, as seen from the fig. above a real, inverted and magnified image A'B' is formed. 
We can see that, $∆ \mathrm{ABO} \mathrm{and} ∆\mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{O} \mathrm{is} \mathrm{similar}$ .

$\therefore$         $\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}} = \frac{OB\text{'}}{OB}$ 

and also, 

$∆\mathrm{A}\text{'}\mathrm{B}\text{'}\mathrm{F} \mathrm{and} ∆\mathrm{OCF} \mathrm{are} \mathrm{similar}.$

              $\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{OC}}=\frac{FB\text{'}}{OF}$ 

But, OC = AB 

$\therefore$          $\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}= \frac{\mathrm{FB}\text{'}}{\mathrm{OF}}$ 
From the above equations, we get 

      $\frac{\mathrm{OB}\text{'}}{\mathrm{OB}} = \frac{FB\text{'}}{OF}= \frac{OB\text{'} - OF}{OF }$
Using the sign convention, 
           OB= -u , OB' = +v, OF = +f 

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

$\Rightarrow$  $\mathrm{vf} = -\mathrm{uv} + \mathrm{uf} \mathrm{or} \mathrm{uv} = \mathrm{uf} -\mathrm{vf}$ 

Dividing both sides by uvf, we have 

          $\frac{\mathrm{uv}}{\mathrm{uvf}} = \frac{uf}{uvf}-\frac{vf}{ uvf}$ 

i.e.,         $\frac{1}{\mathrm{f}} = \frac{1}{v}-\frac{1}{u}$ 

which is the required lens formula. 

ii.) Given, a glass prism of refractive index = 1.5 
When, this prism is immersed in a liquid of refractive index 1.3 then, angle of minimum deviation of the prism reduces.