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

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(A) The shift in the image position due to the glass plate is given by $S = (1 - \frac{1}{\mu}) t = (1 - \frac{1}{1.5}) 	imes 1 \; cm = \frac{1}{3} \

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$3.2$

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(A) The shift in the image position due to the glass plate is given by $S = (1 - \frac{1}{\mu}) t = (1 - \frac{1}{1.5}) 	imes 1 \; cm = \frac{1}{3} \; cm$.First,calculate the focal length $f$ of the lens using the initial conditions: $u = -240 \; cm$ and $v = 12 \; cm$.$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{12} - \frac{1}{-240} = \frac{20 + 1}{240} = \frac{21}{240} \; cm^{-1}$.When the glass plate is introduced,the image must be formed at a distance $v' = 12 - \frac{1}{3} = \frac{35}{3} \; cm$ from the lens so that the plate shifts it back to the film.Using the lens formula again: $\frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} = \frac{3}{35} - \frac{21}{240} = \frac{3}{35} - \frac{7}{80}$.$\frac{1}{u'} = \frac{48 - 49}{560} = -\frac{1}{560} \; cm^{-1}$.Thus,$u' = -560 \; cm = -5.6 \; m$.The shift required for the object is $|u'| - |u| = 5.6 \; m - 2.4 \; m = 3.2 \; m$.

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