The diameter of the aperture of a plano-conve | vedclass

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(C) $1$. Calculate the refractive index $(\mu)$ of the lens material:$\mu = \frac{c}{v} = \frac{3 	imes 10^8 \, m/s}{2 	imes 10^8 \, m/s} = 1.5$.$2$

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$30$

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(C) $1$. Calculate the refractive index $(\mu)$ of the lens material:$\mu = \frac{c}{v} = \frac{3 	imes 10^8 \, m/s}{2 	imes 10^8 \, m/s} = 1.5$.$2$. Find the radius of curvature $(R)$ of the curved surface:Let $a$ be the radius of the aperture,so $a = 3 \, cm = 30 \, mm$. Let $t$ be the maximum thickness,$t = 3 \, mm$.Using the geometric property of a lens: $a^2 = t(2R - t)$.$(30)^2 = 3(2R - 3)$$900 = 6R - 9$$909 = 6R$$R = 151.5 \, mm = 15.15 \, cm$.$3$. Calculate the focal length $(f)$ using the Lens Maker's Formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a plano-convex lens,$R_1 = R = 15.15 \, cm$ and $R_2 = \infty$.$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{15.15} - 0 ight) = 0.5 	imes \frac{1}{15.15} = \frac{1}{30.3}$.Thus,$f \approx 30 \, cm$.

The diameter of the aperture of a plano-convex lens is $6 \, cm$ and its maximum thickness is $3 \, mm$. If the velocity of light in the material of the lens is $2 \times 10^8 \, m/s$,its focal length is ........ $cm$.

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