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

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(A) The focal length $f$ of the lens is given by the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given $v = 12 \ cm$ and $u = -240 \ cm$ $

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$3.2$

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(A) The focal length $f$ of the lens is given by the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given $v = 12 \ cm$ and $u = -240 \ cm$ $(2.4 \ m = 240 \ cm)$.$\frac{1}{f} = \frac{1}{12} - \frac{1}{-240} = \frac{20+1}{240} = \frac{21}{240} \ cm^{-1}$.When a glass plate of thickness $t = 1 \ cm$ and refractive index $\mu = 1.5$ is introduced,the image shifts towards the lens by an amount $\Delta x = t(1 - \frac{1}{\mu}) = 1(1 - \frac{1}{1.5}) = 1(1 - \frac{2}{3}) = \frac{1}{3} \ cm$.The new image position $v' = 12 - \frac{1}{3} = \frac{35}{3} \ cm$.To maintain a sharp focus,the new object distance $u'$ must satisfy: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$.$\frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} = \frac{3}{35} - \frac{21}{240} = \frac{3}{35} - \frac{7}{80}$.$\frac{1}{u'} = \frac{3 	imes 16 - 7 	imes 7}{560} = \frac{48 - 49}{560} = -\frac{1}{560} \ cm^{-1}$.So,$u' = -560 \ cm = -5.6 \ m$.The shift required for the object is $|u'| - |u| = 5.6 \ m - 2.4 \ m = 3.2 \ m$.

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