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An equiconcave lens having radius of curvature of each surface 20 cm has one surface silvered. If the refractive index of the lens is 1.5, then the magnitude of the focal length is

A. 2.5cm
B. 0.4cm
C. 0
D. 5cm

User Takahiro
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1 Answer

6 votes

The magnitude of the focal length of the equiconcave lens with one surface silvered is D. 5 cm.

How to find the magnitude of the focal length?

The focal length (f_single) of a single concave surface of a lens can be calculated using the formula:

f_single = 1 / (2 * (n - 1) x (1 / r))

Plugging in the values, we get:

f_single = 1 / (2 * (1.5 - 1) x (1 / 20))

= 1 / (2 * 0.5 * (1 / 20))

= 10 c

Therefore, the effective focal length (f_effective) of the lens becomes:

f_effective = f_single / 2 = 10 cm / 2

= 5 cm

User Andrew Floren
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