A thin plano-convex lens of focal length 73.5 | vedclass

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(C) For a plano-convex lens, the radius of curvature $R$ is related to the aperture radius $r$ and thickness $t$ by the geometric relation: $R^2 = (R-

Vedclass · Accepted Answer

$1.8 	ext{ mm}$

Vedclass · Answer

(C) For a plano-convex lens, the radius of curvature $R$ is related to the aperture radius $r$ and thickness $t$ by the geometric relation: $R^2 = (R-t)^2 + r^2$. Since $t$ is very small, $R^2 \approx R^2 - 2Rt + t^2 + r^2$, which simplifies to $2Rt \approx r^2$, or $R = \frac{r^2}{2t}$.Given the diameter $d = 8.4 	ext{ cm}$, the radius $r = \frac{d}{2} = 4.2 	ext{ cm}$.Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For a plano-convex lens, $R_1 = R$ and $R_2 = \infty$, so $\frac{1}{f} = (\mu - 1)(\frac{1}{R})$.Thus, $R = f(\mu - 1)$.Substituting $R = \frac{r^2}{2t}$, we get $\frac{r^2}{2t} = f(\mu - 1)$.Rearranging for thickness $t$: $t = \frac{r^2}{2f(\mu - 1)}$.Substituting the values: $r = 4.2 	ext{ cm}$, $f = 73.5 	ext{ cm}$, $\mu = \frac{5}{3}$.$t = \frac{(4.2)^2}{2 	imes 73.5 	imes (\frac{5}{3} - 1)} = \frac{17.64}{147 	imes \frac{2}{3}} = \frac{17.64}{98} = 0.18 	ext{ cm} = 1.8 	ext{ mm}$.

$A$ thin plano-convex lens of focal length $73.5 \text{ cm}$ has a circular aperture of diameter $8.4 \text{ cm}$. If the refractive index of the material of the lens is $\frac{5}{3}$, then the thickness of the lens is nearly

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