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  4. The focal length of a biconvex lens of radii of each surfaces 50 cm and refractive index 1.5, is:

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Q. The focal length of a biconvex lens of radii of each surfaces 50 cm and refractive index 1.5, is:

VMMC MedicalVMMC Medical 2003

Solution:

Using lens makers formula $ \frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right) $ Here, $ {{R}_{1}}=+\,50\,cm,\,{{R}_{2}}=-50cm,\,\mu =1.5 $ $ \frac{1}{f}=(1.5-1)\left( \frac{1}{50}-\frac{1}{(-50)} \right) $ or $ \frac{1}{f}=\frac{1}{50} $ Hence, $ f=50\,cm $