Q.In a displacement-method experiment, a convex lens is moved between a fixed object and a screen kept $90\ \text{cm}$ apart. Two lens positions give a sharp image on the screen, and these two positions are separated by $30\ \text{cm}$. Find the focal length of the lens.

Using the conjugate-foci (displacement) method: the two sharp-image positions are symmetric, giving u1 = (D-x)/2 and v1 = (D+x)/2; the two magnifications are reciprocals, so the ratio of the two image sizes is ((D+x)/(D-x))^2.

Setup: object and screen are a fixed distance D apart. A convex lens between them forms a sharp image at two positions; shifting the lens by x from the first position to the second gives the second sharp image (this displacement is x).

Step 1 - Conjugate-foci symmetry. If at position 1 the object distance is u1 and image distance is v1, then u1 + v1 = D. Because object and image distances are interchangeable for a lens (the ray path is reversible), at position 2 the roles swap: u2 = v1 and v2 = u1.

Step 2 - Use the displacement x. Moving the lens toward the screen by x increases the object distance and decreases the image distance by x, so v1 - u1 = x. Solving with u1 + v1 = D:

u1 = (D - x)/2, v1 = (D + x)/2.

And then u2 = v1 = (D + x)/2, v2 = u1 = (D - x)/2.

Step 3 - Magnifications (size = |m| x object size):

m1 = v1/u1 = (D + x)/(D - x), …