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Question
A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.
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Solution
Given,
Power of the lens (P) = \[\frac{1}{\text{ Focal length}}\] = 5.0 D
The height of the image is four times the height of the object.
i.e. \[h_i = 4 h_0 , \frac{h_i}{h_0} = 4\]
We know magnification (m) is also given by \[m = \frac{h_i}{h_0} = \frac{v}{u} \Rightarrow \frac{v}{u} = 4, v = 4u\]
The lens maker formula is given by \[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
Here, v is the image distance and u is the object distance.
Now,
\[\frac{1}{f} = P = 5 \text{ and } f = \frac{1}{5} m = 20 \text{ cm }\]
\[\frac{1}{4u} - \frac{1}{u} = \frac{1}{20}\]
\[ - \frac{3}{4u} = \frac{1}{20}\]
\[ \Rightarrow u = - \frac{60}{4} = - 15 \text{ cm }\]
Hence, the required object distance is 15 cm.
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