A convex lens focuses an object 20 cm from i | vedclass

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(B) The apparent shift in the position of the image due to the glass slab is given by $d = t(1 - \frac{1}{\mu})$.Substituting the values,$d = 1.4(1 -

Vedclass · Accepted Answer

$30.7$

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(B) The apparent shift in the position of the image due to the glass slab is given by $d = t(1 - \frac{1}{\mu})$.Substituting the values,$d = 1.4(1 - \frac{5}{7}) = 1.4(\frac{2}{7}) = 0.4 \ cm$.First,calculate the focal length $f$ of the lens using the initial conditions: $u = -20 \ cm$ and $v = 5 \ cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{f} = \frac{1}{5} - \frac{1}{-20} = \frac{4+1}{20} = \frac{5}{20} = \frac{1}{4}$. Thus,$f = 4 \ cm$.When the glass slab is inserted,the image must still form on the screen at $v = 5 \ cm$. However,the slab shifts the image position towards the lens by $d = 0.4 \ cm$. Therefore,the effective image distance becomes $v' = 5 - 0.4 = 4.6 \ cm$.Using the lens formula again for the new object distance $u'$: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$.$\frac{1}{4} = \frac{1}{4.6} - \frac{1}{u'} \Rightarrow \frac{1}{u'} = \frac{1}{4.6} - \frac{1}{4} = \frac{4 - 4.6}{18.4} = \frac{-0.6}{18.4}$.$u' = -\frac{18.4}{0.6} \approx -30.66 \ cm \approx 30.7 \ cm$.

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