Two lens of focal lengths -20 cm and +10 cm are put in combination, find the power of the combination.

Question

Two lens of focal lengths -20 cm and +10 cm are put in combination, find the power of the combination. (A) - 1 D (B) - 2 D (C) + 5 D (D) + 2 D

Tardigrade Admin · Accepted Answer

Here f1= - 20 cm, f2= + 10 cm Focal length of combination is. (1/F)=(1/f1)+(1/f2)=(1/-20)+(1/10) (1/F)=(1/20) ⇒ F = 20 cm Power of the combination, P=(100/F) =(100/20)=+5D

Q. Two lens of focal lengths -20 cm and +10 cm are put in combination, find the power of the combination.

- 1 D

- 2 D

+ 5 D

+ 2 D

Here $f_{1} = - 20 c m, f_{2} = + 10 c m$
Focal length of combination is.
$\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{- 20} + \frac{1}{10}$
$\frac{1}{F} = \frac{1}{20}$
$\Rightarrow F = 20 c m$
Power of the combination,
$P = \frac{100}{F}$
$= \frac{100}{20} = + 5 D$

Q. Two lens of focal lengths -20 cm and +10 cm are put in combination, find the power of the combination.

- 1 D

- 2 D

+ 5 D

+ 2 D

Here f1​=−20cm,f2​=+10cm Focal length of combination is. F1​=f1​1​+f2​1​=−201​+101​ F1​=201​ ⇒F=20cm Power of the combination, P=F100​ =20100​=+5D

Here $f_{1} = - 20 c m, f_{2} = + 10 c m$
Focal length of combination is.
$\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{- 20} + \frac{1}{10}$
$\frac{1}{F} = \frac{1}{20}$
$\Rightarrow F = 20 c m$
Power of the combination,
$P = \frac{100}{F}$
$= \frac{100}{20} = + 5 D$