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Numerical Problems · Q6

Q.If the distance DD between an object and a screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called the conjugate foci method. If dd is the distance between the two positions of the lens, obtain the equation for the focal length of the convex lens.

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Step 1. With the object and screen fixed a distance DD apart, and the lens at object distance xx from the object (image distance D−xD-x from the lens), the lens equation gives 1D−x+1x=1f\dfrac{1}{D-x}+\dfrac{1}{x}=\dfrac{1}{f} (writing both distances as positive magnitudes for this real-image, both-on-the-same-side-of-focus setup).

Step 2. Clearing denominators: x+(D−x)=x(D−x)fx+(D-x)=\dfrac{x(D-x)}{f}, i.e. D=x(D−x)fD=\dfrac{x(D-x)}{f}, so x(D−x)=fDx(D-x)=fD, i.e. x2−Dx+fD=0x^2-Dx+fD=0 -- a quadratic in xx with two roots x1,x2x_1,x_2 corresponding to the two lens positions that both form a real image on the screen.

Step 3. By the sum and product of roots of this quadratic, x1+x2=Dx_1+x_2=D and x1x2=fDx_1x_2=fD. The distance between the two lens positions is d=∣x1−x2∣d=|x_1-x_2|, and using the algebraic identity (x1−x2)2=(x1+x2)2−4x1x2(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2: d2=D2−4fDd^2=D^2-4fD. …

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