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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 \mu_{1} and \mu_{2} and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:

  • Option 1)

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

  • Option 2)

    \frac{\text{R}}{2(\mu_{1}+\mu_{2})}

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

best_answer

As we learnt in

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

 

 

Focal length of convex lens is 

\frac{1}{f1}= \left ( \mu _{1} -1 \right ).(\frac{1}{R}-\frac{1}{-R})=\frac{2(\mu -1)}{R}

Focal length of concave lens is 

\frac{1}{f2}= \left ( \mu _{2} -1 \right ).(\frac{1}{-R}-\frac{1}{R})=\frac{2(\mu_{2} -1)}{R}

\therefore Resultant focal length is 

\frac{1}{f}= \frac{1}{f1} +\frac{1}{f2} =\frac{2(\mu_{1} -1)}{R}+\frac{-2(\mu_{2} -1)}{R}

\frac{1}{f}=\frac{2\mu _{1}-\mu _{2}}{R} or F= \frac{R}{2\left ( \mu _{1}- \mu _{2}\right )}


Option 1)

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

This option is incorrect.

Option 2)

\frac{\text{R}}{2(\mu_{1}+\mu_{2})}

This option is incorrect.

Option 3)

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

This option is correct.

Option 4)

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

This option is incorrect.

Posted by

Aadil

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