Curved surfaces of a plano-convex lens of refractive index μ1 and a plano-concave lens of refractive index μ2 have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

Question

Curved surfaces of a plano-convex lens of refractive index μ1 and a plano-concave lens of refractive index μ2 have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses. (A) (1/μ2-μ1) (B) μ1-μ2 (C) (1/μ1-μ2) (D) μ2-μ1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8579323a9fbac998e9ea83664a3ea0db-.png" /> (1/ f 1)=(μ1-1)((1/ R )) (1/ f 2)=(μ2-1)(-(1/ R )) (1/ f 1)+(1/ f 2)=(1/ f eq )=((μ1-1)-(μ2-1)/ R ) (1/ f eq )=((μ1-μ2)/ R ) ( R / f eq )=(μ1-μ2)

Q. Curved surfaces of a plano-convex lens of refractive index $μ_{1}$ and a plano-concave lens of refractive index $μ_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

$\frac{1}{μ _{2} - μ _{1}}$

$μ_{1} - μ_{2}$

$\frac{1}{μ _{1} - μ _{2}}$

$μ_{2} - μ_{1}$

$\frac{1}{f _{1}} = (μ_{1} - 1) (\frac{1}{R})$
$\frac{1}{f _{2}} = (μ_{2} - 1) (- \frac{1}{R})$
$\frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{f _{e q}} = \frac{( μ _{1} - 1 ) - ( μ _{2} - 1 )}{R}$
$\frac{1}{f _{e q}} = \frac{( μ _{1} - μ _{2} )}{R}$
$\frac{R}{f _{e q}} = (μ_{1} - μ_{2})$

Q. Curved surfaces of a plano-convex lens of refractive index μ1​ and a plano-concave lens of refractive index μ2​ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

μ2​−μ1​1​

μ1​−μ2​

μ1​−μ2​1​

μ2​−μ1​