(a) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it. |
(b) In the above diagram mark the object-distance (u) and the image-distance (v) with their proper signs (+ve or ?ve as per the new Cartesians sign convention) and state how these distances are related to the focal length (f) of the concave lens in this case. |
(c) Find the nature and power of a lens which forms a real and inverted image of magnification ?1 at a distance of 40cm from its optical centre. |
Answer:
(a) (b) The object distance (u) and image distance (v) are marked in the diagram of part (a). Relation : \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] (c) As, \[m=-1;\] hence, the lens is convex Now, \[m=\frac{v}{u}\] \[\therefore \] \[v=-u\] Thus, object is at 2F. \[\because \] \[v=40\,cm\] (Given) \[\therefore \] \[2f=40\text{ }cm\] \[f=20\,cm\,=0.2\,m.\] \[P=\frac{1}{f}=\frac{1}{0.2}=+5D\] (convex lens)
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