Question

Find the effective power of the combination of two lenses in contact having powers (i) +5D and +3D (ii) +5D and -3D.

Answer 1

Answer on Question #43896, Physics, Optics

Find the effective power of the combination of two lenses in contact having powers

(i) $+5\mathrm{D}$ and $+3\mathrm{D}$

(ii) $+5\mathrm{D}$ and -3D.

Solution:

Figure 1

In Figure 1, let the focal lengths of the two lenses be $f_1$ and $f_2$ .

$\mathsf{u}_1 = \mathsf{OC}_1$ and $\mathsf{v}_1 = \mathsf{I}_1\mathsf{C}_1$

$\mathsf{u}_2 = -\mathsf{C}_1\mathsf{I}_1$ which is approximately equal to $\mathsf{C}_2\mathsf{I}_1$ and $\mathsf{v}_2 = \mathsf{C}_2\mathsf{I}$ which is approximately equal to $\mathsf{C}_1\mathsf{I}$

Therefore

$$\frac {1}{u _ {1}} + \frac {1}{v _ {1}} = \frac {1}{f _ {1}}$$

and

$$\frac {1}{u _ {2}} + \frac {1}{v _ {2}} = \frac {1}{f _ {2}}$$ $$\frac {1}{O C _ {1}} + \frac {1}{I _ {1} C _ {1}} = \frac {1}{f _ {1}}$$

and

$$\frac {1}{- C _ {1} I _ {1}} + \frac {1}{C _ {1 I}} = \frac {1}{f _ {2}}$$

Therefore:

$$\frac {1}{O C _ {1}} + \frac {1}{I _ {1} C _ {1}} = \frac {1}{f _ {1}} + \frac {1}{f _ {2}} = \frac {1}{F}$$

Combined focal length of two thin lenses in contact is given by:

$$\frac {1}{F} = \frac {1}{f _ {1}} + \frac {1}{f _ {2}}$$

For powers we will have

$$P = P _ {1} + P _ {2}$$

(i) +5D and +3D

$$P = P _ {1} + P _ {2} = 5 D + 3 D = 8 D$$

(ii) +5D and -3D.

$$P = P _ {1} + P _ {2} = 5 D - 3 D = 2 D$$

Answer: (i) $P = 8D$, (ii) $P = 2D$.

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