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(C) Assuming the refractive index of the lens material is $\mu = 1.5$ and the radii of curvature are $R_1 = 20\, cm$ and $R_2 = -20\, cm$,the focal le

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Real,inverted,height $= 4\, cm$

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(C) Assuming the refractive index of the lens material is $\mu = 1.5$ and the radii of curvature are $R_1 = 20\, cm$ and $R_2 = -20\, cm$,the focal length $f$ is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2}) = (1.5 - 1)(\frac{1}{20} - \frac{1}{-20}) = 0.5 	imes \frac{2}{20} = \frac{1}{20}$. Thus,$f = 20\, cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u = -30\, cm$:$\frac{1}{20} = \frac{1}{v} - \frac{1}{-30}$$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$.So,$v = 60\, cm$.The magnification $m$ is given by $m = \frac{v}{u} = \frac{60}{-30} = -2$.Since $m = \frac{h_i}{h_o}$,we have $h_i = m 	imes h_o = -2 	imes 2\, cm = -4\, cm$.The negative sign indicates the image is inverted,and the magnitude of the height is $4\, cm$. Since $v$ is positive,the image is real.

$A$ biconvex lens has a radius of curvature of magnitude $20\, cm$. Which one of the following options best describes the image formed of an object of height $2\, cm$ placed $30\, cm$ from the lens?

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