The image of an object,formed by a plano-conv | vedclass

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(C) Given,image distance $v = 8 \ m$ (real image formed behind the lens).Magnification $m = -\frac{1}{3}$.Using the magnification formula $m = \frac{v

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(C) Given,image distance $v = 8 \ m$ (real image formed behind the lens).Magnification $m = -\frac{1}{3}$.Using the magnification formula $m = \frac{v}{u}$,we get $-\frac{1}{3} = \frac{8}{u}$,which implies $u = -24 \ m$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we have $\frac{1}{f} = \frac{1}{8} - \frac{1}{-24} = \frac{3+1}{24} = \frac{4}{24} = \frac{1}{6}$. Thus,$f = 6 \ m$.The refractive index $\mu$ is given by the ratio of wavelength in vacuum to wavelength in the medium: $\mu = \frac{\lambda_0}{\lambda} = \frac{1}{2/3} = 1.5 = \frac{3}{2}$.Using the Lens Maker's formula for a plano-convex lens: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Here,$R_1 = R$ and $R_2 = -\infty$ (or vice versa),so $\frac{1}{6} = (1.5 - 1) \left( \frac{1}{R} ight) = 0.5 	imes \frac{1}{R} = \frac{1}{2R}$.Therefore,$2R = 6$,which gives $R = 3 \ m$.

The image of an object,formed by a plano-convex lens at a distance of $8 \ m$ behind the lens,is real and is one-third the size of the object. The wavelength of light inside the lens is $\frac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is: (in $m$)

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