An object of height 25 cm is placed in front | vedclass

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(B) Given: $h_1 = 25 \ cm$,$f = 30 \ cm$,$h_2 = \pm 50 \ cm$.Case $1$: Real image $(h_2 = -50 \ cm)$Magnification $m = \frac{h_2}{h_1} = \frac{-50}{25

Vedclass · Accepted Answer

$15$

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(B) Given: $h_1 = 25 \ cm$,$f = 30 \ cm$,$h_2 = \pm 50 \ cm$.Case $1$: Real image $(h_2 = -50 \ cm)$Magnification $m = \frac{h_2}{h_1} = \frac{-50}{25} = -2$.Using $m = \frac{f}{f + u}$,we get $-2 = \frac{30}{30 + u} \Rightarrow 30 + u = -15 \Rightarrow u = -45 \ cm$.Using $m = \frac{v}{u}$,we get $-2 = \frac{v}{-45} \Rightarrow v = 90 \ cm$.The distance between the object and the image is $d = |v| + |u| = 90 + 45 = 135 \ cm$.Case $2$: Virtual image $(h_2 = +50 \ cm)$Magnification $m = \frac{h_2}{h_1} = \frac{50}{25} = +2$.Using $m = \frac{f}{f + u}$,we get $2 = \frac{30}{30 + u} \Rightarrow 60 + 2u = 30 \Rightarrow 2u = -30 \Rightarrow u = -15 \ cm$.Using $m = \frac{v}{u}$,we get $2 = \frac{v}{-15} \Rightarrow v = -30 \ cm$.The distance between the object and the image is $d = |v - u| = |-30 - (-15)| = |-15| = 15 \ cm$.Since $15 \ cm$ is one of the options,the correct answer is $15 \ cm$.

An object of height $25 \ cm$ is placed in front of a convex lens of focal length $30 \ cm$. If the height of the formed image is $50 \ cm$,then the distance between the object and the image is ..... $cm$.

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