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  • A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index  of the first lens

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surface are of the same radius of curvature \mathrm{R = 14\ cm}. For this biconvex lens, for an object distance of \mathrm{40\ cm}, the image distance will be

Option: 1

\mathrm{-280.0\ cm}


Option: 2

\mathrm{40.0\ cm}


Option: 3

\mathrm{21.5\ cm}


Option: 4

\mathrm{13.3\ cm}


Answers (1)

best_answer

The focal length \mathrm{\left ( f_{1} \right )} of the lens with \mathrm{n = 1.5} is given by
\mathrm{\frac{1}{\mathrm{f}_1}=\left(\mathrm{n}_1-1\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]=(1.5-1)\left[\frac{1}{14}-\frac{1}{\infty}\right]=\frac{1}{28}}
The focal length \mathrm{\left ( f_{2} \right )} of the lens with \mathrm{n = 1.2} is given by
\mathrm{\frac{1}{\mathrm{f}_2}=\left(\mathrm{n}_2-1\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]=(1.2-1)\left[\frac{1}{\infty}-\frac{1}{-14}\right]=\frac{1}{70}}
The focal length \mathrm{F} of the combination is \mathrm{\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}=\frac{1}{20}}
Applying lens formula for the combination of lens
\frac{1}{V}-\frac{1}{U}=\frac{1}{F} \Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20} \Rightarrow V=40 \mathrm{~cm}

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Divya Prakash Singh

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