An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is then

1 m

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0.5 m

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0.24 m

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2 m

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Solution

The correct option is C 0.24 m
Here, $m = \frac{v}{u} = - 4$
$\Rightarrow u = \frac{- v}{4}$ ------------(1)
Also, $| u | + | v | = 1.5$
$\frac{v}{4} + v = 1.5$
$\Rightarrow \frac{(v + 4 v)}{4} = 1.5$
$\Rightarrow v = 1.2 m$
So, putting the value of $v$ in equation (1):
$u = \frac{- 1.2}{4} = - 0.3 m$
$∴ f = \frac{u v}{u - v}$
$\Rightarrow f = \frac{(- 0.3 \times 1.2)}{(- 0.3 - 1.2)} = 0.24 m$
Hence the correct option is $(C)$

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An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is then

The correct option is C 0.24 mHere, m=vu=−4⇒u=−v4------------(1)Also, |u|+|v|=1.5v4+v=1.5⇒(v+4v)4=1.5 ⇒v=1.2m So, putting the value of v in equation (1): u=−1.24=−0.3m∴f=uvu−v⇒f=(−0.3×1.2)(−0.3−1.2)=0.24mHence the correct option is (C)