(i) A convex lens is placed on an optical bench and is moved till it gives a real image of an object at minimum distance of 80 cm from the latter. Find the focal length of the lens. If object is placed at a distance of 15 cm from the lens, find the position of image. |
(ii) A double convex lens is made of glass of refractive index 1.5 with both faces of the same radius of curvature. |
Find the radius of curvature required, if the focal length is 15 cm. |
Answer:
(i) The minimum distance between an object and its real image formed by a convex lens is 4f \[\therefore \] \[4f=80\] or \[f=\frac{80}{4}=20\,cm\] Now \[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}=\frac{1}{15}=\frac{1}{60}\] \[\therefore \] \[v=-60\,cm\] (ii) According to lens maker's formula, \[\frac{1}{f}=(\mu -1)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] Given, \[{{R}_{1}}=+R\] and \[{{R}_{2}}=-R,\] \[f=15\,cm,\] \[\mu =1.5\] \[\frac{1}{f}=(1.5-1)\,\left( \frac{1}{R}-\frac{1}{(-R)} \right);\]\[\frac{1}{0.15}=0.5\times \frac{2}{R}\] \[\therefore \] \[R==0.15\,m=15\,cm\]
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