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A double convex lens has two surfaces of equal radii R and refractive index $μ = 1.5$ , we have

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$f = R/2$

$f = - R$

$f = 2 R$

$f = R$

answer is B.

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Detailed Solution

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$μ = 1.5, R_{1} = {R,R}_{2} = - R,f = + f$

$\frac{1}{f} = (1.5 - 1) (\frac{1}{R} + \frac{1}{R}) \Rightarrow \frac{1}{f} = 0.5 \times \frac{2}{R} = \frac{1}{2} \times \frac{2}{R}$

$\frac{1}{f} = \frac{1}{R} \Rightarrow f = + R$

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