In the displacement method of a lens,we obtai | vedclass

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(C) In the displacement method,let $D$ be the distance between the object and the screen,and $d$ be the distance between the two positions of the lens

Vedclass · Accepted Answer

$\frac{md}{m^2-1}$

Vedclass · Answer

(C) In the displacement method,let $D$ be the distance between the object and the screen,and $d$ be the distance between the two positions of the lens. The focal length $f$ is given by the formula $f = \frac{D^2 - d^2}{4D}$. For the two positions of the lens,the magnifications are $m_1 = m$ and $m_2 = 1/m$. Using the relation $m = \frac{v}{u}$,we know that for the two positions,$v_1 = u_2$ and $u_1 = v_2$. Also,$D = u_1 + v_1$ and $d = v_1 - u_1$. From these,$v_1 = \frac{D+d}{2}$ and $u_1 = \frac{D-d}{2}$. Since $m = \frac{v_1}{u_1} = \frac{D+d}{D-d}$,we can solve for $D$ in terms of $m$ and $d$: $m(D-d) = D+d \implies D(m-1) = d(m+1) \implies D = d \frac{m+1}{m-1}$. Substituting $D$ into the focal length formula: $f = \frac{D^2 - d^2}{4D} = \frac{d^2 [(\frac{m+1}{m-1})^2 - 1]}{4d(\frac{m+1}{m-1})} = \frac{d [\frac{(m+1)^2 - (m-1)^2}{(m-1)^2}]}{4(\frac{m+1}{m-1})} = \frac{d [\frac{4m}{(m-1)^2}]}{4(\frac{m+1}{m-1})} = \frac{md}{m^2-1}$.

In the displacement method of a lens,we obtain two images for different positions of the lens. One image is magnified,and the other is diminished. If $m$ is the magnification of the first image,then the focal length of the lens is given by:

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