Question

Two thin lenses of focal lengthh +20cm and -30cm are placed in contact the effective power of combination is?

Answer 1

Answer

The effective power of the combination is +1.67 Dioptre

Explanation

Let us consider

the focal length of the two lenses be f and f'

the focal length of the combination be F

The power of the combination of lenses be P

Formula to be used ( since the lense are in contact )

☆1/F = P

☆1/F = 1/f1 + 1/f2 + 1/f3+............1/fn

Given data

f = +20cm

f' = -30

Therefore,

P = 1/F

→P = 1/f + 1/f'

→P = 1/20cm + 1/(-30cm)

→ P = 100/20m - 100/30m

→ P = 5D - 3.33D

→ P = +1.67D

Answer 2

Answer:

$\large\boxed{\sf{+1.67\;D}}$

Explanation:

Given that there are two thin lenses.

Lst the focal length of lenses be f1 and f2 respectively.

According to question, we have,

• f1 = +20 cm
• f2 = -30 cm

Now, it's said that these are placed in contact to each other.

To find the effective power of combination.

We know that,

When n number of lenses are kept in contact, having focal length f1, f2, f3,..........fn

Then, the reciprocal of focal length of combination is the sum of reciprocals of all the respective focal lengths.

Now, let the focal length of combination is F.

Therefore, we will get,

$= > \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ \\ = > \frac{1}{F} = \frac{1}{20} + \frac{1}{ - 30} \\ \\ = > \frac{1}{F} = \frac{1}{20} - \frac{1}{30} \\ \\ = > \frac{1}{F} = \frac{30 - 20}{30 \times 20} \\ \\ = > \frac{1}{F} = \frac{10}{600} \\ \\ = > \frac{1}{F} = \frac{1}{60} \\ \\ = > F= 60$

.°.Focal length of combination is 60 cm or 0.6 m.

Now, to find the power, P.

We know that, Power is the reciprocal of focal length.

Therefore, we will get,

$= > P = \frac{1}{0.6} \\ \\ = > P = \frac{10}{6} \\ \\ = > P = \frac{5}{3} \\ \\ = > P = + 1.67$

Hence, the power of combination is +1.67 D.