Question

A biconvex lens has focal length f and intensity of light I passing through it. What will be the focal length and intensity for portions of lenses obtained by cutting it vertically and horizontally as shown in the figure?

Answer 1

R₁ = R

R₂ = - R

Vertically-

${\displaystyle \frac{1}{\text{f}}=(\mu -1)(\frac{1}{{\left({\text{R}}_1\right)}_{\to \text{R}}}-\frac{1}{{\left({\text{R}}_2\right)}_{\to -\text{R}}})}$

${\displaystyle \frac{1}{\text{f}}=(\mu -1)\frac{2}{\text{R}}}$   ...(1)

${\displaystyle \frac{1}{{\text{f}}_{\text{p}}}=(\mu -1)(\frac{1}{{\infty }_{\to 0}}-(-\frac{1}{\text{R}}))}$

${\displaystyle \frac{1}{{\text{f}}_{\text{p}}}=\frac{\mu -1}{\text{R}}}$   ....(2)

equation (1) ÷ equation (2)

${\displaystyle \frac{\frac{1}{\text{f}}}{\frac{1}{{\text{f}}_{\text{p}}}}}$ = 2

${\displaystyle \frac{{\text{f}}_{\text{p}}}{\text{f}}}$ = 2

${\displaystyle {\text{f}}_{\text{p}}}$ = 2f

Horizontally-

${\displaystyle \frac{1}{\text{f'}}=(\mu -1)(\frac{1}{{\left({\text{R}}_1\right)}_{\to \text{R}}}-\frac{1}{{\left({\text{R}}_2\right)}_{\to -\text{R}}})}$

${\displaystyle \frac{1}{\text{f'}}=(\mu -1)\frac{2}{\text{R}}}$   ...(3)

f' = f