Question

A convex lens of focal length 80 cm and a
concave lens of focal length 50 cm are combined
together. What will be their resulting power?
(a) + 6.5 D (b) – 6.5 D
(c) + 7.5 D (d) – 0.75 D

Answer 1

$\huge{\red{Answer}}$

$\boxed{\huge{ Option\: (d) \: -0.75 \: D}}$

Given:

Focal length of Convex Lens, f1 = 80 cm

Focal length of Concave lens, f2 = - 50 cm

Solution:

We know,

Power of combination (P) = P1 +P2 =

$= \frac{100}{f1} + \frac{100}{f2} \\ \\ = \frac{100}{80} + \frac{100}{ - 50} \\ \\ = \frac{100}{80} - \frac{100}{50} \\ \\ = - 0.75$

So the resulting power will be -0.75 D (d)

Answer 2

AnswEr :

-0.75D

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

A convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together.

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

Their resulting power.

$\bf{\orange{\large{\underline{\underline{\bf{Explanation\::}}}}}}$

$\bf{We\:have}\begin{cases}\sf{Focal\:length\:of\:convex\:lenses\:(F_{1})=80\:cm}\\ \sf{Focal\:length\:of\:concave\:lenses\:(F_{2})=-50\:cm}\end{cases}}$

Formula use : (combination of focal length);

$\sf{\purple{\large{\bf{\dfrac{1}{F} =\dfrac{1}{F_{1}} +\dfrac{1}{F_{2}} }}}}$

$\leadsto\tt{\dfrac{1}{F} =\dfrac{1}{F_{1}} +\dfrac{1}{F_{2}} }\\\\\\\\\leadsto\tt{\dfrac{1}{F} =\dfrac{1}{80} +\bigg(\dfrac{1}{-50} \bigg)}}\\\\\\\\\leadsto\tt{\dfrac{1}{F} =\dfrac{1}{80} -\dfrac{1}{50} \:\:\:\:\:\: \underbrace{\sf{Taking\:L.C.M.\:of\:80\:\&\:50}}}}\\\\\\\\\leadsto\tt{\dfrac{1}{F} =\dfrac{5-8}{400} }\\\\\\\\\leadsto\tt{\dfrac{1}{F} =\dfrac{-3}{400} }\\\\\\\\\leadsto\tt{\red{F=\dfrac{400}{-3} cm}}$

$\large{\dag{\sf{\underline{\underline{\sf{Convert\:into\:metres(m)\::}}}}}}}}$

$\leadsto\tt{F=\dfrac{400}{\frac{-3}{100} }}\\\\\\\\\leadsto\tt{F=\dfrac{4\cancel{00}}{-3} \times \dfrac{1}{\cancel{100}}} \\\\\\\\\leadsto\tt{\boxed{\purple{\tt{F=-\dfrac{4}{3} m}}}}}$

$\large{\dag{\sf{\underline{\underline{\sf{Formula\:use\:(Power\:of\:lens)\::}}}}}}}}$

$\longmapsto\sf{Power\:of\:lens\:(P)=\dfrac{1}{Focal\:length\:of\:the\:lens} }\\\\\\\\\longmapsto\sf{Power\:of\:lens\:(P)=-\cancel{\dfrac{3}{4}} }\\\\\\\\\longmapsto\sf{\purple{Power\:of\:lens\:(P)=-0.75\:D}}$

∴ The resulting power of lens is -0.75D.

We know that S.I. unit of power of lens is Dioptre (D).