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(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 \[{{R}_{2}}=-R,\] \[f=15\,cm,\] \[\mu =1.5\] \[\frac{1}{f}=(1.5-1)\,\left( \frac{1}{R}-\frac{1}{(-R)} \right);\]\[\frac{1}{0.15}=0.5\times \frac{2}{R}\] \[\therefore \]      \[R==0.15\,m=15\,cm\]