CUET Preparation Today
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$\text{Let radius of two surfaces are R and 2R respectively}$ $ \frac{1}{F} = (\mu -1) (\frac{1}{R_1} - \frac{1}{R_2}) = (1.5 -1) (\frac{1}{R} - \frac{1}{-2R}) = \frac{3}{4R}$ $\Rightarrow F = \frac{4R}{3}$ $\Rightarrow R = \frac{3F}{4} = 4.5 cm$ $\text{larger radius of curvature is } 2R = 9cm$ |
