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

The correct option is B Rμ_2 μ_1 1F = 1f_1 + 1f_2 = 1 μ_1R + μ_2 1R 1F=1 μ_1+μ_2 1R=μ_2 μ_1R F=Rμ_2 μ_1 Option C ...

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Byju's Answer

Standard XII

Physics

Power and Thin Lenses in Contact

A plano-conve...

Question

A plano-convex lens (focal length $f_{2}$ , refractive index $μ_{2}$ , radius of curvature $R$ ) fits exactly into a plano-concave lens (focal length $f_{1}$ , 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 :

f_{1} - f_{2}

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f_{1} + f_{2}

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\frac{R}{μ_{2} - μ_{1}}

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\frac{2 f_{1} f_{2}}{f_{1} + f_{2}}

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Solution

The correct option is B $\frac{R}{μ_{2} - μ_{1}}$
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1 - μ_{1}}{R} + \frac{μ_{2} - 1}{R}$

$\frac{1}{F} = \frac{1 - μ_{1} + μ_{2} - 1}{R} = \frac{μ_{2} - μ_{1}}{R}$

$F = \frac{R}{μ_{2} - μ_{1}}$
Option C

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

The correct option is B Rμ2−μ11F=1f1+1f2=1−μ1R+μ2−1R1F=1−μ1+μ2−1R=μ2−μ1RF=Rμ2−μ1Option C

The correct option is B $\frac{R}{μ_{2} - μ_{1}}$
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1 - μ_{1}}{R} + \frac{μ_{2} - 1}{R}$

$\frac{1}{F} = \frac{1 - μ_{1} + μ_{2} - 1}{R} = \frac{μ_{2} - μ_{1}}{R}$

$F = \frac{R}{μ_{2} - μ_{1}}$
Option C