Question

Two lenses of focal length 20cm and minus 40 cm are kept in contact with each other the power of equivalent lens wiki

Answer 1

f1=20cm

P1= 100!20=5D

f2= -40cm

P2=-100!40=-2.5D

equivalent P=5-2.5=+2.5D

Answer 2

The power of equivalent lens is 7.5 D.

Explanation:

Given that,

The focal length of the lens 1, $f_1=20\ cm$

The focal length of the lens 2, $f_2=40\ cm$

We need to find the power of equivalent lens. It is given by :

$P=P_1+P_2$

Since,

$P=\dfrac{1}{f}$

$\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$

$\dfrac{1}{f}=\dfrac{1}{20}+\dfrac{1}{40}$

$\dfrac{1}{f}=0.075\ m^{-1}$

$\dfrac{1}{f}=7.5\ cm^{-1}$

i.e. $P=7.5\ D$

So, the power of equivalent lens is 7.5 D. Hence, this is the required solution.

Learn more,

Power

https://brainly.in/question/7665148