For an object placed at a distance 2.4,m from | vedclass

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(B) $1$. Applying the lens formula for the initial state:$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$Given $v = 12\,cm = 0.12\,m$ and $u = -2.4\,m$:$\fra

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$3.2$

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(B) $1$. Applying the lens formula for the initial state:$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$Given $v = 12\,cm = 0.12\,m$ and $u = -2.4\,m$:$\frac{1}{0.12} - \frac{1}{-2.4} = \frac{1}{f} \Rightarrow \frac{100}{12} + \frac{1}{2.4} = \frac{1}{f} \Rightarrow \frac{25}{3} + \frac{1}{2.4} = \frac{20+1}{2.4} = \frac{21}{2.4} = \frac{1}{f} \Rightarrow f = \frac{2.4}{21} = \frac{0.8}{7}\,m$.$2$. When a glass slab of thickness $t$ and refractive index $\mu$ is introduced,the image shifts by $\Delta x = t(1 - \frac{1}{\mu})$.$\Delta x = 1\,cm 	imes (1 - \frac{1}{1.5}) = 1\,cm 	imes (1 - \frac{2}{3}) = \frac{1}{3}\,cm$.$3$. The new image position $v'$ required to be on the screen is $v' = 12\,cm - \frac{1}{3}\,cm = \frac{35}{3}\,cm = \frac{35}{300}\,m = \frac{7}{60}\,m$.$4$. Applying the lens formula again for the new object distance $u'$:$\frac{1}{v'} - \frac{1}{u'} = \frac{1}{f}$$\frac{60}{7} - \frac{1}{u'} = \frac{21}{2.4} = \frac{210}{24} = \frac{35}{4}$$\frac{1}{u'} = \frac{60}{7} - \frac{35}{4} = \frac{240 - 245}{28} = -\frac{5}{28}$$u' = -\frac{28}{5} = -5.6\,m$.$5$. The shift in the object position is $|u' - u| = |-5.6 - (-2.4)| = |-3.2| = 3.2\,m$.

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