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(A) Let the object distance be $u_1$ and image distance be $v_1$. For a real image,the magnification is $m_1 = -v_1/u_1$. Since the image is on a scre

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

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(A) Let the object distance be $u_1$ and image distance be $v_1$. For a real image,the magnification is $m_1 = -v_1/u_1$. Since the image is on a screen,$v_1$ is positive. Let $v_1 = q$. Then $u_1 = -q/m_1$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{q} - \frac{1}{-q/m_1} = \frac{1}{f} \Rightarrow \frac{1+m_1}{q} = \frac{1}{f} \Rightarrow q = f(1+m_1)$.When the lens is moved by distance $x$,the new image distance is $v_2 = q+x$. The new magnification is $m_2 = -v_2/u_2$. Thus $u_2 = -v_2/m_2 = -(q+x)/m_2$.Using the lens formula again:$\frac{1}{q+x} - \frac{1}{-(q+x)/m_2} = \frac{1}{f} \Rightarrow \frac{1+m_2}{q+x} = \frac{1}{f} \Rightarrow q+x = f(1+m_2)$.Subtracting the two expressions for $q$ and $q+x$:$(q+x) - q = f(1+m_2) - f(1+m_1)$$x = f(m_2 - m_1)$$f = \frac{x}{m_2 - m_1}$.

$A$ convex lens forming a real image of magnification $m_1$ on a screen is moved through a distance $x$. $A$ new image of magnification $m_2$ is again formed on the screen. The focal length of the lens is:

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