A convex lens of focal length F is placed in contact with a concave lens of focal length F- The equivalent focal length of the combination is-Infinity-F-2-2FZero

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Toppr Admin · Accepted Answer

Given that-xA0-f1-F -amp- f2-x2212-FLet the equivalent focal length be f-x21D2-1f-1f1-1f2-xA0- -xA0- -xA0- -xA0- -xA0- -xA0-1F-1-x2212-F-xA0- -xA0- -xA0- -xA0- -xA0- -xA0-0Therefore- -xA0-f-10-x221E-

A convex lens of focal length $F$ is placed in contact with a concave lens of focal length $F$ . The equivalent focal length of the combination is:
Infinity
$F / 2$
$2 F$
Zero

Given that:
$f_{1} = F$ & $f_{2} = - F$

Let the equivalent focal length be $f$ .

$\Rightarrow \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

$= \frac{1}{F} + \frac{1}{- F}$

$= 0$

Therefore, $f = \frac{1}{0} = \infty$

A convex lens of focal length F is placed in contact with a concave lens of focal length F. The equivalent focal length of the combination is:InfinityF/22FZero

Given that: f1=F & f2=−FLet the equivalent focal length be f.⇒1f=1f1+1f2 =1F+1−F =0Therefore, f=10=∞

A convex lens of focal length $F$ is placed in contact with a concave lens of focal length $F$ . The equivalent focal length of the combination is:
Infinity
$F / 2$
$2 F$
Zero

Given that:
$f_{1} = F$ & $f_{2} = - F$

Let the equivalent focal length be $f$ .

$\Rightarrow \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

$= \frac{1}{F} + \frac{1}{- F}$

$= 0$

Therefore, $f = \frac{1}{0} = \infty$