The radii of curvature of the faces of a doub | vedclass

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(B) Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given:$f = +12\, cm$$R_1 = +10\, cm$$R_2 = -1

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given:$f = +12\, cm$$R_1 = +10\, cm$$R_2 = -15\, cm$Substituting the values:$\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-15} \right)$$\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{15} \right)$$\frac{1}{12} = (\mu - 1) \left( \frac{3 + 2}{30} \right)$$\frac{1}{12} = (\mu - 1) \left( \frac{5}{30} \right)$$\frac{1}{12} = (\mu - 1) \left( \frac{1}{6} \right)$$\mu - 1 = \frac{6}{12} = 0.5$$\mu = 1 + 0.5 = 1.5$

The radii of curvature of the faces of a double convex lens are $10\, cm$ and $15\, cm$. Its focal length is $12\, cm$. What is the refractive index of the glass?

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