A biconvex lens is made of material of refractive index 1.5 each of radius of curvature of lens surface is 20 cm. The focal length of lens is:

20 cm

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10 cm

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25 cm

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30 cm

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Solution

The correct option is A 20 cm
From Lens maker's formula,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$r_{1} = 20 c m, R_{2} = - R_{1} = - 20 c m$

$∴ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20})$
$= 0.5 (\frac{1}{10})$ or $f = 20 c m$

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A biconvex lens of a focal length of 10 cm is to be made from a glass material. If the refractive index of the material is 1.5, what must be the radius of curvature of the surface of the lens?

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A biconvex lens is made of material of refractive index 1.5 each of radius of curvature of lens surface is 20 cm. The focal length of lens is:

The correct option is A 20 cmFrom Lens maker's formula,1f=(μ−1)(1R1−1R2)r1=20cm,R2=−R1=−20cm∴1f=(1.5−1)(120+120)=0.5(110) or f=20cm

The correct option is A 20 cm
From Lens maker's formula,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$r_{1} = 20 c m, R_{2} = - R_{1} = - 20 c m$

$∴ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20})$
$= 0.5 (\frac{1}{10})$ or $f = 20 c m$