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A plano-convex lens (focal length f₂ , refractive index $\mu_{2}$ , radius of curvature R ) fits exactly into a plano-concave lens ( focal length f₁ , refractive index $\mu_{1}$ , radius of curvature R ) . Their plane surfaces are parallel to each other. Then , the focal length of the combination will be :
Option 1)
$\frac{R}{\mu_{2} - \mu_{1} }$
Option 2)
f₁ + f₂
Option 3)
f₁ - f₂
Option 4)
$\frac{2f_{1}f_{2}}{f_{1} + f_{2}}$

Answers (1)

best_answer

Lensmaker's Formula -

$\frac{1}{f}= \left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}- \frac{1}{R_{2}}\right )$

- wherein

$\mu _{1}=$ refractive index of medium of object

$\mu _{2}=$ refractive index of lens

$R_{1}\, and \, R_{2}$ are radius of curvature of two surface

$\frac{1}{f_{eq}} = \frac{1}{f_{1}}+\frac{1}{f_{2}}$

and

$\frac{1}{f_{1}} = \left ( \mu _{2}-1 \right )\left ( \frac{1}{\infty } -\frac{1}{R}\right )$

$\frac{1}{f_{2}} = \left ( \mu _{2}-1 \right )\left ( \frac{1}{R } -\frac{1}{\infty}\right )$

$\Rightarrow \frac{1}{f_{eq}} = -\frac{\mu _{1}-1}{R}+\frac{\mu _{2}-1}{R}$

$\frac{1}{f_{eq}} = \frac{R}{\mu _{2}-\mu _{1}}$

Option 1)
$\frac{R}{\mu_{2} - \mu_{1} }$
Option 2)

f₁ + f₂

Option 3)

f₁ - f₂

Option 4)

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

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