An object 2.4 , m in front of a lens forms a | vedclass

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(A) According to the thin lens formula:$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$Given $u = -2.4 \, m = -240 \, cm$ and $v = 12 \, cm$.$\frac{1}{f} = \

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$5.6$

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(A) According to the thin lens formula:$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$Given $u = -2.4 \, m = -240 \, cm$ and $v = 12 \, cm$.$\frac{1}{f} = \frac{1}{12} - \frac{1}{-240} = \frac{20 + 1}{240} = \frac{21}{240} \, cm^{-1}$.When a glass plate of thickness $t = 1 \, cm$ and refractive index $\mu = 1.5$ is introduced,the shift produced is:$	ext{Shift} = t \left(1 - \frac{1}{\mu}ight) = 1 \left(1 - \frac{1}{1.5}ight) = 1 \left(1 - \frac{2}{3}ight) = \frac{1}{3} \, cm$.To keep the image on the film,the lens must now form the image at a distance $v' = 12 - \frac{1}{3} = \frac{35}{3} \, cm$.Using the lens formula again with the same focal length:$\frac{21}{240} = \frac{1}{v'} - \frac{1}{u'} = \frac{3}{35} - \frac{1}{u'}$.$\frac{1}{u'} = \frac{3}{35} - \frac{21}{240} = \frac{3}{35} - \frac{7}{80} = \frac{48 - 49}{560} = -\frac{1}{560} \, cm^{-1}$.$u' = -560 \, cm = -5.6 \, m$.The object should be shifted to a distance of $5.6 \, m$ from the lens.

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