What is the refractive index of the material | vedclass

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(A) Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given: $f = \frac{20}{3} \ cm$, $R_1 = 5 \ cm

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$1.5$

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(A) Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Given: $f = \frac{20}{3} \ cm$, $R_1 = 5 \ cm$, $R_2 = -10 \ cm$ (for a double convex lens).Substituting the values:$\frac{3}{20} = (\mu - 1) \left( \frac{1}{5} - \frac{1}{-10} ight)$$\frac{3}{20} = (\mu - 1) \left( \frac{2 + 1}{10} ight)$$\frac{3}{20} = (\mu - 1) \left( \frac{3}{10} ight)$$\mu - 1 = \frac{3}{20} 	imes \frac{10}{3}$$\mu - 1 = \frac{1}{2}$$\mu = 1 + 0.5 = 1.5$

What is the refractive index of the material of a double convex lens having radii of curvature of $5 \ cm$ and $10 \ cm$ and focal length of $\frac{20}{3} \ cm$?

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