A double convex lens made of glass has both radii of curvature of magnitude 20 cm. Incident light rays parallel to the axis of the lens will converge at a distance at
[Given refractive index of glass = 1.5 ]

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\frac{40}{3}

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Solution

The correct option is A 20
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Given, $μ = 15, R_{1} = + 20 c m, R_{2} = - 20 c m$
$\frac{1}{f} = (15 - 1) (\frac{1}{20} + \frac{1}{20}) = 0.5 \times \frac{2}{20} = \frac{1}{20}$
$f = 20 c m$
Incident light lays will convenge at its second principle focus.

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A double convex lens made of glass has both radii of curvature of magnitude 20 cm. Incident light rays parallel to the axis of the lens will converge at a distance at[Given refractive index of glass = 1.5 ]

The correct option is A 201f=(μ−1)(1R1−1R2)Given, μ=15,R1=+20cm,R2=−20cm1f=(15−1)(120+120)=0.5×220=120f=20cmIncident light lays will convenge at its second principle focus.

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