The refractive index of the material of a dou | vedclass

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a double convex lens,the radii of

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$5.0$

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a double convex lens,the radii of curvature are $R_1 = R$ and $R_2 = -R$.Given $f = 5 ~cm$ and $\mu = 1.5$.Substituting these values into the formula:$\frac{1}{5} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} ight)$$\frac{1}{5} = 0.5 	imes \left( \frac{1}{R} + \frac{1}{R} ight)$$\frac{1}{5} = 0.5 	imes \frac{2}{R}$$\frac{1}{5} = \frac{1}{R}$Therefore,$R = 5 ~cm$.

The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 ~cm$. If the radii of curvature are equal,the value of the radius of curvature (in $cm$) is

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