A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ1 and μ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of combination is

Question

A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ1 and μ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of combination is (A)  (2R/(μ2 - μ1)) (B)  (R/2(μ1 + μ2)) (C)  (R/2(μ1 - μ2)) (D)  (R/(μ1 - μ2))

Tardigrade Admin · Accepted Answer

Equivalent focal length is given by (1/feq) = (1/f1) + (1/f2) (1/feq) = (μ1- 1)((1/∞) - (1/-R))+ (μ2 -1)((1/-R) - (1/∞)) ⇒ feq = (R/μ1 - μ2)

$\frac{2 R}{( μ _{2} - μ _{1} )}$

$\frac{R}{2 ( μ _{1} + μ _{2} )}$

$\frac{R}{2 ( μ _{1} - μ _{2} )}$

$\frac{R}{( μ _{1} - μ _{2} )}$

Equivalent focal length is given by $\frac{1}{f _{e q}} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$
$\frac{1}{f _{e q}} = (μ_{1} - 1) (\frac{1}{\infty} - \frac{1}{- R}) + (μ_{2} - 1) (\frac{1}{- R} - \frac{1}{\infty})$
$\Rightarrow f_{e q} = \frac{R}{μ _{1} - μ _{2}}$

(μ2​−μ1​)2R​

2(μ1​+μ2​)R​

2(μ1​−μ2​)R​

(μ1​−μ2​)R​

Equivalent focal length is given by feq​1​=f1​1​+f2​1​ feq​1​=(μ1​−1)(∞1​−−R1​)+(μ2​−1)(−R1​−∞1​) ⇒feq​=μ1​−μ2​R​

$\frac{2 R}{( μ _{2} - μ _{1} )}$

$\frac{R}{2 ( μ _{1} + μ _{2} )}$

$\frac{R}{2 ( μ _{1} - μ _{2} )}$

$\frac{R}{( μ _{1} - μ _{2} )}$

Equivalent focal length is given by $\frac{1}{f _{e q}} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$
$\frac{1}{f _{e q}} = (μ_{1} - 1) (\frac{1}{\infty} - \frac{1}{- R}) + (μ_{2} - 1) (\frac{1}{- R} - \frac{1}{\infty})$
$\Rightarrow f_{e q} = \frac{R}{μ _{1} - μ _{2}}$