A planoconvex lens has a maximum thickness of | vedclass

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(D) Case $1$: Curved surface in contact with the table.The real depth of the lens is $t = 6 \,cm$. The apparent depth is $d' = 4 \,cm$.The refractive

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$34$

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(D) Case $1$: Curved surface in contact with the table.The real depth of the lens is $t = 6 \,cm$. The apparent depth is $d' = 4 \,cm$.The refractive index $n$ is given by $n = \frac{	ext{Real depth}}{	ext{Apparent depth}} = \frac{6}{4} = 1.5$.Case $2$: Plane surface in contact with the table.We use the refraction formula at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here, the light travels from the lens $(n = 1.5)$ to the air $(n = 1)$.Real depth $u = 6 \,cm$, apparent depth $v = -\frac{17}{4} \,cm$ (virtual image).Using the formula $\frac{1}{v} - \frac{1.5}{u} = \frac{1 - 1.5}{R}$:$\frac{1}{-17/4} - \frac{1.5}{6} = \frac{-0.5}{R}$$-\frac{4}{17} - 0.25 = -\frac{0.5}{R}$$-\frac{4}{17} - \frac{1}{4} = -\frac{1}{2R}$$-\frac{16 + 17}{68} = -\frac{1}{2R}$$\frac{33}{68} = \frac{1}{2R} \implies R = \frac{68}{66} 	imes 33 = 34 \,cm$.

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