A point source is located at a distance of 20 | vedclass

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(A) Using the lens maker's formula for a biconvex lens,we have:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Since the lens is

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$\frac{20}{3} \ cm$

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(A) Using the lens maker's formula for a biconvex lens,we have:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Since the lens is symmetrical and biconvex,$R_1 = 5 \ cm$ and $R_2 = -5 \ cm$.$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{5} - \frac{1}{-5} ight) = 0.5 	imes \left( \frac{2}{5} ight) = \frac{1}{5} \ cm^{-1}$.Thus,the focal length $f = 5 \ cm$.Using the lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -20 \ cm$:$\frac{1}{v} - \frac{1}{-20} = \frac{1}{5}$$\frac{1}{v} = \frac{1}{5} - \frac{1}{20} = \frac{4 - 1}{20} = \frac{3}{20}$$v = \frac{20}{3} \ cm$.The image is formed at a distance of $\frac{20}{3} \ cm$ from the optical center of the lens.

$A$ point source is located at a distance of $20 \ cm$ from the front surface of a symmetrical glass biconvex lens with equal radii of curvature $5 \ cm$. The distance at which the image is formed from the rear surface of this lens is $[$Given refractive index of the glass is $1.5]$

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