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(B) In the displacement method,for a fixed distance $D$ between the object and the screen,there are two positions of the lens where a sharp image is f

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

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(B) In the displacement method,for a fixed distance $D$ between the object and the screen,there are two positions of the lens where a sharp image is formed.Let the object distance be $u$ and image distance be $v$ for the first position. Then $m_1 = v/u$.For the second position,by the principle of reversibility,the object distance becomes $v$ and the image distance becomes $u$. Thus,$m_2 = u/v$.Note that $m_1 	imes m_2 = (v/u) 	imes (u/v) = 1$.Given the displacement of the lens is $x = v - u$.We know $m_1 - m_2 = v/u - u/v = (v^2 - u^2) / (uv) = (v - u)(v + u) / (uv) = x(v + u) / (uv)$.From the lens formula,$1/f = 1/v - 1/(-u) = (u + v) / (uv)$.Substituting this into the expression for $m_1 - m_2$,we get $m_1 - m_2 = x 	imes (1/f) = x/f$.Therefore,the focal length $f = x / (m_1 - m_2)$.

In the displacement method,a convex lens is placed between an object and a screen. If the magnifications in the two positions are $m_1$ and $m_2$ and the displacement of the lens between the two positions is $x$,then the focal length of the lens is:

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