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  • A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is <img alt="\mathrm{4D}" src="https://entrancecorner.onco

A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is \mathrm{4D}, then the power of a cut lens will be:

Option: 1

\mathrm{2D}


Option: 2

\mathrm{3D}


Option: 3

\mathrm{4D}


Option: 4

\mathrm{5D}


Answers (1)

Biconvex lens is cut perpendicular to the principal axis, it will become a plano-convex lens. For length of biconvex lens,

\begin{aligned} & \frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & \frac{1}{\mathrm{f}}=(\mathrm{n}-1) \frac{2}{\mathrm{R}} \quad\left(\because \mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=-\mathrm{R}\right) \\ & \Rightarrow \mathrm{f}=\frac{\mathrm{R}}{2}(\mathrm{n}-1)\ \ ................\mathrm{(i)} \end{aligned}
For plano-convex lens, 
\begin{aligned} & \frac{1}{\mathrm{f}_1}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right) \\ & \mathrm{f}_1=\frac{\mathrm{R}}{(\mathrm{n}-1)}\ \ \ \ ..........\mathrm{(ii)} \end{aligned}
On comparing eqs. \mathrm{(i)} and \mathrm{(ii)}, we see the focal length become double.
As power of lens, \mathrm{\mathrm{P} \propto \frac{1}{\text { focal length }}} 
Hence, power will become half. 
New power \mathrm{=\frac{4}{2}=2D}
???????

Posted by

Kshitij

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