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Question
A double convex lens of + 5 D is made of glass of refractive index 1.55 with both faces of equal radii of curvature. Find the value of its radius of curvature.
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Solution
From lens maker formula, we have
\[P = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]
\[\text { where} \]
\[ P =\text { Power of lens } = + 5 D\]
\[\mu = \text { Refractive index of the lens } = 1 . 55\]
\[ R_1 =\text { Radius of curvature of first face } \left( + ve \right)\]
\[ R_2 = \text { Radius of curvature of second face } \left( - ve \right)\]
\[\text {Given }: \]
\[ R_1 = R_2 = R\]
\[ \Rightarrow P = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]
\[ \Rightarrow 5 = (1 . 55 - 1)\left( \frac{1}{R} - \frac{1}{- R} \right)\]
\[ \Rightarrow 5 = (1 . 55 - 1)\left( \frac{2}{R} \right)\]
\[ \Rightarrow 5 = 0 . 55\left( \frac{2}{R} \right)\]
\[ \Rightarrow R = \frac{0 . 55 \times 2}{5}\]
\[ \Rightarrow R = 0 . 22 m\]
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