A convex lens is placed between an object and | vedclass

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Question

(B) Let the distance between the object and the screen be $D$. For the displacement method,the two positions of the lens are separated by distance $d$

Vedclass · Accepted Answer

$\frac{d}{m_1 - m_2}$

Vedclass · Answer

(B) Let the distance between the object and the screen be $D$. For the displacement method,the two positions of the lens are separated by distance $d$.For the first position,magnification $m_1 = \frac{v_1}{u_1}$. Since $v_1 + u_1 = D$ and $v_1 - u_1 = d$,we have $v_1 = \frac{D+d}{2}$ and $u_1 = \frac{D-d}{2}$. Thus,$m_1 = \frac{D+d}{D-d}$.For the second position,the object and image distances are interchanged,so $m_2 = \frac{D-d}{D+d}$.Note that $m_1 \cdot m_2 = 1$,so $m_2 = \frac{1}{m_1}$.The displacement $d$ is related to focal length $f$ by $f = \frac{D^2 - d^2}{4D}$.Using $m_1 - m_2 = \frac{D+d}{D-d} - \frac{D-d}{D+d} = \frac{(D+d)^2 - (D-d)^2}{D^2 - d^2} = \frac{4Dd}{D^2 - d^2}$.Since $f = \frac{D^2 - d^2}{4D}$,we have $m_1 - m_2 = \frac{d}{f}$.Therefore,$f = \frac{d}{m_1 - m_2}$.

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