A double convex lens made of glass has both r | vedclass

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(D) 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 radius

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$20 \,cm$

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(D) 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 radius of curvature of the first surface $R_1$ is positive $(+20 \,cm)$ and the radius of curvature of the second surface $R_2$ is negative $(-20 \,cm)$ according to the sign convention.Given: $\mu = 1.5$, $R_1 = 20 \,cm$, $R_2 = -20 \,cm$.Substituting these values into the formula:$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} ight)$$\frac{1}{f} = (0.5) \left( \frac{1}{20} + \frac{1}{20} ight)$$\frac{1}{f} = 0.5 	imes \frac{2}{20} = 0.5 	imes 0.1 = 0.05$$f = \frac{1}{0.05} = 20 \,cm$.Since the incident rays are parallel to the axis, they converge at the focal point, which is at a distance $L = f = 20 \,cm$ from the pole.

$A$ double convex lens made of glass has both radii of curvature of magnitude $20 \,cm$. Incident light rays parallel to the axis of the lens will converge to a point at a distance $L$ from the common pole $P$. The value of $L$ is [Refractive index of glass $= 1.5$].

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