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Question
A convex lens has a focal length of 10 cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of (a) 9.8 cm, (b) 10.2 cm from the lens.
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Solution
Given:
Focal length (f) of the convex lens = 10 cm
(a) As per the question, the object distance (u) is 9.8 cm.
The lens equation is given by:
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
\[\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{9 . 8}\]
= \[\frac{1}{v} = \frac{9 . 8 - 10}{98} = \frac{- 0 . 2}{98}\]
= v = − 98 × 5
= − 490 cm (Same side of the object)
v = 490 cm (Virtual and on on the side of object)
Magnification of the image
= `v/u`
\[= \frac{- 490}{- 9 . 8}\]
\[ = 50\]
Therefore, the image is erect and virtual.
(b) Object distance, u = 10.2 cm
The lens equation is given by:
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
= \[\frac{1}{v} = \frac{1}{10} - \frac{1}{10 . 2}\]
\[= \frac{10 . 2 - 10}{102} = \frac{0 . 2}{102}\]
= v = 102 × 5 = 510 cm (Real and on the opposite side of the object)
Magnification of the image \[= \frac{v}{u}\]
\[= \frac{510}{- 9 . 8}\]
\[ = - 52 . 04\]
Therefore, the image is real and inverted. 
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