An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.
CBSE DELHI - SET 1 2013
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We have height of object, $\text{h}_1 = 6 \text{ cm}$, focal length of lens, f = -5 cm and object distance, u = - 10 cm Using lens formula, we have $\frac{1}{v}-\frac{1}{u}= \frac{1}{f}$$\Rightarrow\frac{1}{v}-\frac{1}{(-10)}=\frac{1}{(-5)} \Rightarrow\frac{1}{v}+\frac{1}{10}= -\frac{1}{5}\Rightarrow\frac{1}{v}= -\frac{1}{5}-\frac{1}{10}$
$\Rightarrow v= -\frac{10}{3}= -3.33 \text{ cm}$
$\text{Magnification, M} = \frac{v}{u} = -\frac{10}{3}\times\bigg(-\frac{1}{10}\bigg)=\frac{1}{3}$
$\text{Again, Magnification, M} = \frac{v}{u}= \frac{\text{h}_{2}}{\text{h}_{1}} \Rightarrow \frac{\text{h}_{2}}{6}= \frac{1}{3}\Rightarrow \text{h}_{2} = \frac{6}{3} = 2 \text{ cm }$
Thus the image will be formed in front of the lens at a distance of 3.33 cm from the lens, virtual and erect of size 2 cm.
$\Rightarrow v= -\frac{10}{3}= -3.33 \text{ cm}$
$\text{Magnification, M} = \frac{v}{u} = -\frac{10}{3}\times\bigg(-\frac{1}{10}\bigg)=\frac{1}{3}$
$\text{Again, Magnification, M} = \frac{v}{u}= \frac{\text{h}_{2}}{\text{h}_{1}} \Rightarrow \frac{\text{h}_{2}}{6}= \frac{1}{3}\Rightarrow \text{h}_{2} = \frac{6}{3} = 2 \text{ cm }$
Thus the image will be formed in front of the lens at a distance of 3.33 cm from the lens, virtual and erect of size 2 cm.
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