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 the thickness $t$ by the geometry: $R^2 = r^2 + (R-

Vedclass · Accepted Answer

$1.8 \ mm$

Vedclass · Answer

(C) For a plano-convex lens, the radius of curvature $R$ is related to the aperture radius $r$ and the thickness $t$ by the geometry: $R^2 = r^2 + (R-t)^2$. Since $t$ is very small, $R^2 = r^2 + R^2 - 2Rt + t^2 \approx r^2 + R^2 - 2Rt$, which simplifies to $R = \frac{r^2}{2t}$.Using the Lens Maker's Formula: $\frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$. For a plano-convex lens, $R_1 = R$ and $R_2 = \infty$, so $\frac{1}{f} = (n-1) \frac{1}{R}$, which gives $R = f(n-1)$.Equating the two expressions for $R$: $f(n-1) = \frac{r^2}{2t}$.Given $f = 73.5 \ cm$, $n = \frac{5}{3}$, and diameter $d = 8.4 \ cm$, so $r = \frac{d}{2} = 4.2 \ cm$.Substituting the values: $73.5 	imes \left(\frac{5}{3} - 1ight) = \frac{(4.2)^2}{2t}$.$73.5 	imes \frac{2}{3} = \frac{17.64}{2t} \Rightarrow 49 = \frac{8.82}{t}$.$t = \frac{8.82}{49} \ cm = 0.18 \ cm = 1.8 \ mm$.

$A$ thin plano-convex lens of focal length $73.5 \ cm$ has a circular aperture of diameter $8.4 \ cm$. 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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