A glass convex lens is of refractive index $1 \cdot 55$ with both faces of same radius of curvature. What will be the radius of curvature if focal length is to be $20 \mathrm{~cm}$ ?

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Aicharya · Accepted Answer

Lens maker's formula, $\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$ Here, $\mathrm{f}=20 \mathrm{~cm}, \mu=1.55, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=-\mathrm{R}$ $\begin{aligned} &\frac{1}{20}=(1.55-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \text { or } \frac{1}{20}=0.55 \times \frac{2}{R} \ &\Rightarrow R=1.1 \times 20=22 \mathrm{~cm} \end{aligned}$ .

A glass convex lens is of refractive index $1 \cdot 55$ with both faces of same radius of curvature. What…

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