(a) Define focal length of a divergent lens. |
(b) A divergent lens of focal length 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed. |
(c) Draw a ray diagram to show the formation of image in the above situation. |
Answer:
(a) The point from which rap of light parallel to principal axis after refraction, appear to diverge is called principal focus of a divergent lens and the distance between optical centre and this focus is called focal length of a divergent lens. (b) Given: \[f=-30\,cm,\,u=?,\,v=-15\,cm,\,{{h}_{o}}=6\,cm,\,{{h}_{1}}=?\] We know that, \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] or \[\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\] \[\Rightarrow \] \[u=\frac{vf}{f-v}=\frac{-15\,\times -30}{-30+15}\] \[=\frac{450}{-15}=-30cm\] Now, \[m=\frac{{{h}_{I}}}{{{h}_{O}}}=\frac{v}{u}\] \[\Rightarrow \] \[{{h}_{I}}=\frac{v}{u}\times \,{{h}_{o}}=\frac{-15}{-30}\times 6=3cm\] (c)
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