A thin biconvex lens, constructed from a material with refractive index μ_2 , is flanked by a medium of refractive index μ_1 on one side and a medium of refractive index μ_3 on the opposite side. Both radii of curvature

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Correct answer: B. (μ_3 R)/(2μ_2 - μ_1 - μ_3)

Applying the formula for refraction at a spherical surface for the first surface:

\dfrac{\mu_2}{v_1} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R_1}

Given that the incident beam is parallel to the principal axis,

u = -\infty

. For a biconvex lens,

R_1 = +R

.

\dfrac{\mu_2}{v_1} = \dfrac{\mu_2 - \mu_1}{R}

For the second surface, the image formed by the first surface acts as a virtual object.

\dfrac{\mu_3}{v} - \dfrac{\mu_2}{v_1} = \dfrac{\mu_3 - \mu_2}{R_2}

For the biconvex lens,

R_2 = -R

.

\dfrac{\mu_3}{v} - \dfrac{\mu_2}{v_1} = \dfrac{\mu_3 - \mu_2}{-R} = \dfrac{\mu_2 - \mu_3}{R}

Adding the equations for both surfaces:

\dfrac{\mu_3}{v} = \dfrac{\mu_2 - \mu_1}{R} + \dfrac{\mu_2 - \mu_3}{R}

\dfrac{\mu_3}{v} = \dfrac{2\mu_2 - \mu_1 - \mu_3}{R}

v = \dfrac{\mu_3 R}{2\mu_2 - \mu_1 - \mu_3}

A thin biconvex lens, constructed from a material with refractive...

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