The distance between an object and its image | vedclass

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(B) Given,magnification $m = 2$ and the image is real. For a real image,the distance between the object and the image is $D = |u| + |v| = 0.72 \ m$.Si

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given,magnification $m = 2$ and the image is real. For a real image,the distance between the object and the image is $D = |u| + |v| = 0.72 \ m$.Since $m = \frac{|v|}{|u|} = 2$,we have $|v| = 2|u|$.Substituting this into the distance equation: $|u| + 2|u| = 0.72 \ m \Rightarrow 3|u| = 0.72 \ m \Rightarrow |u| = 0.24 \ m$ and $|v| = 0.48 \ m$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $v = 0.48 \ m$ and $u = -0.24 \ m$:$\frac{1}{f} = \frac{1}{0.48} - \frac{1}{-0.24} = \frac{1 + 2}{0.48} = \frac{3}{0.48} \Rightarrow f = 0.16 \ m$.When the object is moved $0.04 \ m$ towards the lens,the new object distance is $u' = -(0.24 - 0.04) = -0.20 \ m$.Using the lens formula again: $\frac{1}{v'} - \frac{1}{-0.20} = \frac{1}{0.16} \Rightarrow \frac{1}{v'} = \frac{1}{0.16} - \frac{1}{0.20} = \frac{5 - 4}{0.80} = \frac{1}{0.80} \Rightarrow v' = 0.80 \ m$.The new magnification is $m' = \frac{v'}{u'} = \frac{0.80}{-(-0.20)} = 4$.

The distance between an object and its image produced by a converging lens is $0.72 \ m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04 \ m$ towards the lens?

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