Discussion Forum

 The equi- conves lens has a focal length 'f' . If the lens is cut along the line perpendicular to the principal  axis and passing through the pole, what will be the focal length of any half part ?

Answer:

Explanation:

 Key idea:     The lens Maker's formula is given as 

$\frac{1}{f}= (\mu_{med}-1)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$

 where , f= focal length of lens, R₁= radius of first curved part and R₂= Radius of second curved part 

 As, for equiconvex lens R₁= R₂= R (say)

 So,  $\frac{1}{f}= (\mu_{real}-1)\frac{2}{R_{}}$   ......(I)

Now, if lens  is cut along the line perpendicular to the prinicipal axis as shown in the figure

,

 The new cut  of the lens has, R₁= R and R₂=  $\infty$ . Again by using the lens Maker's formula, focal length of the new part of the lens,

$\frac{1}{f'}= (\mu_{real}-1)\left[\frac{1}{R_{}}-\left(-\frac{1}{\infty}\right)\right]$

$\Rightarrow\frac{1}{f}= (\mu_{real}-1)\left[\frac{1}{R_{}}\right]$   .....(ii)

 So, from the Eqs.(i) and (ii) , we get

  f'=2f