Question

Two lenses are in contacta converging one of focal length 30 cm
and a diverging one of focal length -10 cm. What is the local
length and power of the combination?​

Answer 1

Answer:

Given:

2 lens are in contact. The lens are :

• Converging lens of 30 cm focal length
• Diverging Length of -10 cm focal length.

They are placed in contact with one another.

To find:

Focal length and power of combination.

Concept:

For a combination of lens , the net focal length is given as follows :

$\dfrac{1}{fnet} = \dfrac{1}{f1} + \dfrac{1}{f2} + ...n \: times$

Calculation:

Since we have 2 lens , we will use the above relationship and be careful about the sign of focal length .

$\dfrac{1}{fnet} = \dfrac{1}{30} + \dfrac{1}{ (- 10)}$

$= > \dfrac{1}{fnet} = \dfrac{1 - 3}{30}$

$= > \dfrac{1}{fnet} = \dfrac{ - 2}{30}$

$= > \dfrac{1}{fnet} = \dfrac{ - 1}{15}$

So focal length of combination is -15 cm.

We can say that the combination acts as a concave lens.

For power Calculation :

$\boxed{power = \dfrac{1}{f(in \: metres)}}$

$= > power = \dfrac{1}{0.15}$

$= > power \: = 6.67 \: diopters$

Answer 2

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