The distance between a real object and its real image is 56 cm formed by converging lens, focal length of lens is

f \leq 14

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f > 14

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f = 14

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f = 28

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Solution

The correct option is C $f = 14$
Sol" |u|+|v|=56
$\frac{1}{f} = \frac{1}{| v |} + \frac{1}{| μ |}$
or, $f = \frac{| u | | v |}{| u | + | v |} = \frac{| μ | | v |}{56}$
$| u | | v |$ is maximum when $| u | = | v | = 28 c m$
$\Rightarrow f_{max} = 14$
$f = 14 c m$ en
Nate: Carect oftion is (C)

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The distance between a real object and its real image is 56 cm formed by converging lens, focal length of lens is

The correct option is C f=14Sol" |u|+|v|=561f=1|v|+1|μ|or, f=|u||v||u|+|v|=|μ||v|56|u||v| is maximum when |u|=|v|=28 cm⇒fmax=14f=14cm enNate: Carect oftion is (C)

The correct option is C $f = 14$
Sol" |u|+|v|=56
$\frac{1}{f} = \frac{1}{| v |} + \frac{1}{| μ |}$
or, $f = \frac{| u | | v |}{| u | + | v |} = \frac{| μ | | v |}{56}$
$| u | | v |$ is maximum when $| u | = | v | = 28 c m$
$\Rightarrow f_{max} = 14$
$f = 14 c m$ en
Nate: Carect oftion is (C)