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Lens Maker's Formula

The effective...

Question

The effective focal length of Huygens eyepiece is 3cm. The focal power of field lens is

A

6D

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B

16.7D

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C

3D

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D

10D

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Solution

The correct option is B 16.7D
Effective focal length, $F = 3 c m$
For Huygens eyepiece, if focal length of field lens is f, then effective focal length $F = \frac{f}{2}$

=> $f = 2 F = 6 c m = 0.06 m$

Focal power of field lens = $\frac{1}{f}$ = $16.7 D$

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