A glass convex lens is of refractive index 1. | vedclass

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(A) According to the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given that the lens is a convex lens w

Vedclass · Accepted Answer

$22$

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(A) According to the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Given that the lens is a convex lens with both faces having the same radius of curvature,we have $R_1 = R$ and $R_2 = -R$.The focal length $f = 20 	ext{ cm}$ and refractive index $\mu = 1.55$.Substituting these values into the formula:$\frac{1}{20} = (1.55 - 1) \left( \frac{1}{R} - \frac{1}{-R} ight)$$\frac{1}{20} = 0.55 	imes \left( \frac{1}{R} + \frac{1}{R} ight)$$\frac{1}{20} = 0.55 	imes \frac{2}{R}$$\frac{1}{20} = \frac{1.1}{R}$$R = 1.1 	imes 20 = 22 	ext{ cm}$Therefore,the radius of curvature is $22 	ext{ cm}$.

$A$ glass convex lens is of refractive index $1.55$ with both faces of same radius of curvature. What will be the radius of curvature if the focal length is to be $20 \text{ cm}$ (in $\text{ cm}$)?

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