Question

The focal length of lens I is $+5cm$, lens II is $+10cm$ and lens III is $+20cm$. The total power of the combination of the lens is.

• A. $+0.1D$
• B. $+0.35D$
• C. $+0.75D$
• D. $–0.35D$

Answer 1

The correct option is B $+0.35D$

Given,
Focal length of lens $I=+5cm$
Focal length of lens $II=+10cm$
Focal length of lens $III=+20cm$
We know, Power of lens $P=\frac{1}{\text{focal length}}$.
Total Power = ${\text{P}}_{|}$ + ${\text{P}}_{||}$ + ${\text{P}}_{|||}$ ; P = $\frac{1}{\text{f}}$
So, ${\text{P}}_{|}$ = $\frac{1}{5}$ D ; ${\text{P}}_{||}$ = $\frac{1}{10}$ D ; ${\text{P}}_{|||}$ = $\frac{1}{20}$ D
Total Power of combination ${\text{P}}_{\text{t}}={\text{P}}_{|}$ + ${\text{P}}_{||}$ + ${\text{P}}_{|||}$ ;
${\text{P}}_{\text{t}}=\frac{1}{5}$ + $\frac{1}{10}$ + $\frac{1}{20}$ = $\frac{7}{20}$
${\text{P}}_{\text{t}}$ = $+0.35D$.
Therefore, the total power of combination is $+0.35D$.