The radius of the curvature of the convex surface of a plano-convex lens is $20$ cm. If the refractive index of the material of the lens is $1.5$ . Its focal length is:

10 c m

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20 c m

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30 c m

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40 c m

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Solution

The correct option is D $40 c m$
From Lens makers formula, $\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$R_{1} = \infty$

$R_{2} = - 20 c m$

$\frac{1}{f} = (1.5 - 1) (\frac{1}{20})$

$f = 40 c m$

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The radius of the curvature of the convex surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens is 1.5. Its focal length is:

The correct option is D 40cmFrom Lens makers formula, 1f=(μ−1)(1R1−1R2)R1=∞R2=−20 cm1f=(1.5−1)(120)f=40cm

The radius of the curvature of the convex surface of a plano-convex lens is $20$ cm. If the refractive index of the material of the lens is $1.5$ . Its focal length is:

The correct option is D $40 c m$
From Lens makers formula, $\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$R_{1} = \infty$

$R_{2} = - 20 c m$

$\frac{1}{f} = (1.5 - 1) (\frac{1}{20})$

$f = 40 c m$