A double convex lens made of glass $(μ = 1.5)$ has both radii of curvature of magnitude 20 cm. Incident rays parallel to the axis of the lens will converge at a distance L such that:

L = 20 cm

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L = 10 cm

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L = 40 cm

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L = \frac{20}{3} c m

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Solution

The correct option is A L = 20 cm
When rays incident parallel to lens axis then it will converge at the focus and vice versa $F = d, R_{1} = + 20 c m, R_{2} = - 20 c m$
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}] \frac{1}{L} = (1.5 - 1) [\frac{1}{20} - \frac{1}{- 20}] L + 20 c m$

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A double convex lens made of glass (μ=1.5) has both radii of curvature of magnitude 20 cm. Incident rays parallel to the axis of the lens will converge at a distance L such that:

The correct option is A L = 20 cmWhen rays incident parallel to lens axis then it will converge at the focus and vice versa F=d,R1=+20cm,R2=–20cm 1f=(μ−1)[1R1−1R2]1L=(1.5−1)[120−1−20]L+20cm

A double convex lens made of glass $(μ = 1.5)$ has both radii of curvature of magnitude 20 cm. Incident rays parallel to the axis of the lens will converge at a distance L such that:

The correct option is A L = 20 cm
When rays incident parallel to lens axis then it will converge at the focus and vice versa $F = d, R_{1} = + 20 c m, R_{2} = - 20 c m$
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}] \frac{1}{L} = (1.5 - 1) [\frac{1}{20} - \frac{1}{- 20}] L + 20 c m$