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Home » A double convex lens has focal length . The radius of curvature of one of the surfaces is double of the other. Find the radii if the refractive index of the material of the lens is 1.5. (1) (2) (3) (4)
A double convex lens has focal length 25 \mathrm{~cm}. The radius of curvature of one of the surfaces is double of the other. Find the radii if the refractive index of the material of the lens is 1.5. (1) 50 \mathrm{~cm}, 100 \mathrm{~cm} (2) 100 \mathrm{~cm}, 50 \mathrm{~cm} (3) 25 \mathrm{~cm}, 50 \mathrm{~cm} (4) 18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}
Physics | Physics
A double convex lens has focal length 25 \mathrm{~cm}. The radius of curvature of one of the surfaces is double of the other. Find the radii if the refractive index of the material of the lens is 1.5. (1) 50 \mathrm{~cm}, 100 \mathrm{~cm} (2) 100 \mathrm{~cm}, 50 \mathrm{~cm} (3) 25 \mathrm{~cm}, 50 \mathrm{~cm} (4) 18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}

Answer (4)
Sol. Focal length of lens is \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),

\begin{array}{l} \frac{1}{25}=(1.5-1)\left(\frac{1}{R}+\frac{1}{2 R}\right) \\ \frac{1}{25}=0.5\left(\frac{3}{2 R}\right) \\ 2 R=37.5 \mathrm{~cm} \\ R=18.75 \mathrm{~cm} \end{array}
Therefore, radii are 18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}

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