Question

The focal length of concave lens is 20cm.the focal length of convex lens is 25cm.these two are placed in contact with each other. is it diverging Or converging in nature? ​

Answer 1

Given,

the focal length of convex lens ($f_{1}$) = 20cm

the focal length of concave lens($f_{2}$) = -20cm

We know,

The combination of focal length (f) =$\frac{1}{f} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

$\frac{1}{f}=\frac{1}{25}+\frac{1}{-20} \\\\=\frac{1}{-100} \\\\f=-100cm\\\\=-1.0m$

The lens system a diverging lens of focal length 1.0 m.

Answer 2

AnswEr :

$\bf{\blue{\underline{\underline{\bf{Given\::}}}}}$

The focal length of concave lens is 20 cm.The focal length of convex lens is 25 cm. these two are placed in contact with each other.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

It is diverging or converging in nature.

$\bf{\green{\underline{\underline{\bf{Explanation\::}}}}}$

$\bf{We\:have}\begin{cases}\sf{The\:focal\:length\:of\:concave\:lens\:(f_{2})=-20cm}\\ \sf{The\:focal\:length\:of\:convex\:lens\:(f_{1})=+25}\end{cases}}$

A/q

Formula use : (Combination of focal length)

$\bf{\large{\boxed{\sf{\frac{1}{F} =\frac{1}{F_{1}} +\frac{1}{F_{2}} }}}}}$

$\leadsto\tt{\dfrac{1}{F} =\dfrac{1}{F_{1}} +\dfrac{1}{F_{2}} }\\\\\\\\\leadsto\tt{F=\dfrac{F_{1}F_{2}}{F_{2}+F_{1}} }\\\\\\\\\leadsto\tt{F=\dfrac{25\times (-20)}{-20+25} }\\\\\\\\\leadsto\tt{F=\cancel{\dfrac{-500}{5} }}\\\\\\\\\leadsto\tt{\red{F=-100\:cm}}$

∴ The system behaves as the diverging lens of focal length