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One plano-convex and one plano-concave lens of same radius of curvature 'R' but of

different materials are joined side by side as shown in the figure. If the refractive index of the material

of 1 is $\mu _{1}$ and that of 2 is $\mu _{2}$ , then the focal length of the combination is:

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

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

Option 3)
$\frac{R}{2(\mu_{1}-\mu_{2})}$

Option 4)
$\frac{R}{2-(\mu_{1}-\mu_{2})}$

Answers (1)

S solutionqc

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

Lens placed close to each other -

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

$\frac{1}{f_{eq}}=$ Equivalent focal length.

- wherein

$f_{1},f_{2},------f_{n}$ are focal lenght of lenc 1, 2, 3, -----n

For lens 1

$\frac{1}{f_1}=(\mu _1-1)(\frac{1}{\infty }-\frac{1}{-R})$ ..................(1)

For lens 2

$\frac{1}{f_2}=(\mu _2-1)(\frac{1}{-R}-\frac{1}{\infty })$ ..................(2)

$\frac{1}{f_eq}=\frac{1}{f_1}+\frac{1}{f_2}$

$=\frac{\mu _1-1}{R}+\frac{1-\mu _2}{R}$

$f_{eq}=\frac{R}{\mu _1-\mu _2}$

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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