A plano-convex lens (focal length f2, refractive index μ 2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index μ 1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be: (a) f1 + f2 (b) f1 - f2 (c) (R)/(μ₂ - μ₁) (d) (2 f₁ f₂)/(f₁ + f₂)

Answer: (c). Plano-convex ( μ₂): 1/f₂ = (μ₂-1)/(R); plano-concave ( μ₁): 1/f₁ = -(μ₁-1)/(R). 1/F = 1/f₁ + 1/f₂ = (μ₂-μ₁)/(R) So the answer is (c) (R)/(μ₂-μ₁).

A plano-convex lens (focal length f2, refractive index μ 2, radius of… | JEE Main 2019 (12 Jan, Shift 2)

Q59JEE Main 2019 · 12 JAN · SHIFT 2Medium

Combination of Lenses

A plano-convex lens (focal length f2, refractive index $\mu$ 2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index $\mu$ 1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

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Answer: (c)

Plano-convex ( $\mu_2$ ): $\frac{1}{f_2} = \frac{\mu_2-1}{R}$ ; plano-concave ( $\mu_1$ ): $\frac{1}{f_1} = -\frac{\mu_1-1}{R}$ .
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{\mu_2-\mu_1}{R}$
So the answer is (c) $\frac{R}{\mu_2-\mu_1}$ .

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Combination of Lenses question from JEE Main 2019 (12 Jan, Shift 2)

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