The refractive index of a diverging meniscus lens is 1.5 and its radii of curvatures are 20 cm, 25 cm. The focal power of the lens is:

+0.5 D

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-0.5 D

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+2 D

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-2 D

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Solution

The correct option is C -0.5 D
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$= (1.5 - 1) (\frac{1}{- 20} - \frac{1}{- 25})$
$\frac{1}{f} = - \frac{1}{200}$
$f = - 200 c m$
$p = \frac{100}{f}$
$p = - 0.5 D$

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The refractive index of a diverging meniscus lens is 1.5 and its radii of curvatures are 20 cm, 25 cm. The focal power of the lens is:

The correct option is C -0.5 D1f=(μ−1)(1R1−1R2) =(1.5−1)(1−20−1−25)1f=−1200f=−200 cmp=100fp=−0.5 D

The correct option is C -0.5 D
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$= (1.5 - 1) (\frac{1}{- 20} - \frac{1}{- 25})$
$\frac{1}{f} = - \frac{1}{200}$
$f = - 200 c m$
$p = \frac{100}{f}$
$p = - 0.5 D$