(a) Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus. |
(b) In the above ray diagram mark the object-distance (u) and the image-distance (v) with their proper signs (+ve or ?ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the convex lens in this case. |
(c) Find the power of a convex lens which forms a real, and inverted image of magnification ?1 of an object placed at a distance of 20 cm from its optical centre. |
Answer:
(a) (b) The object distance and image distance are marked in the diagram of part (a). The relation between v, u and f is given by the formula, \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] (c) Given, \[m=-1;\,u=-20\,cm;\,v=?,\,f=?\] We know that, \[m=\frac{v}{u}\] \[-1=\frac{v}{-20}\] \[v=+\,20\,cm\] Thus object is at 2F. i.e., \[2f=20\text{ }cm\] \[f=\frac{20}{2}=10cm=0.1m\] \[P=\frac{1}{f}=\frac{1}{0.1}=+10D\] The power of convex lens is +10 D.
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