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# Engineering

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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