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An equiconvex lens of focal length 10 cm is made up of material with refractive index 1.5. The radius of curvature of the each surface is |
10 cm 20 cm 40 cm 5 cm |
10 cm |
The correct answer is Option (1) → 10 cm Lens maker's formula $\frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$ $\frac{1}{f}=0.5\left[\frac{2}{R}\right]$ R = 10 cm |
