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

For plano convex lens : 1f_1=μ_1 1R For plano concave lens : 1f_2=μ_2 1 R The equivalent focal length of the combination is : 1f_eq=1f_1+1f_2=(μ_1 1) (μ_2 1)R ...

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Standard X

Physics

Lens Formula

Curved surfac...

Question

Curved surfaces of a plano-convex lens of refractive index $\mu_1$ and a plano-concave lens of refractive index $\mu _2$ have equal radius of curvatures as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

μ_{2} - μ_{1}

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μ_{1} - μ_{2}

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\frac{1}{μ_{2} - μ_{1}}

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\frac{1}{μ_{1} - μ_{2}}

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Solution

The correct option is B $μ_{1} - μ_{2}$

For plano convex lens :
$\frac{1}{f_{1}} = \frac{μ_{1} - 1}{R}$
For plano concave lens :
$\frac{1}{f_{2}} = \frac{μ_{2} - 1}{- R}$
The equivalent focal length of the combination is :
$\frac{1}{f_{e q}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{(μ_{1} - 1) - (μ_{2} - 1)}{R}$
$\frac{1}{f_{e q}} = \frac{μ_{1} - μ_{2}}{R}$
$\Rightarrow \frac{R}{f_{e q}} = μ_{1} - μ_{2}$

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Curved surfaces of a plano-convex lens of refractive index \(\mu_1\) and a plano-concave lens of refractive index \(\mu _2\) have equal radius of curvatures as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

The correct option is B μ1−μ2 For plano convex lens : 1f1=μ1−1R For plano concave lens : 1f2=μ2−1−R The equivalent focal length of the combination is : 1feq=1f1+1f2=(μ1−1)−(μ2−1)R 1feq=μ1−μ2R ⇒Rfeq=μ1−μ2