Using the Thin-lens equation A 1f = 1p + 1q. with equation B L =.pdf

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Using the Thin-lens equation: A 1/f = 1/p + 1/q. with equation: B L = p + q and q = p + d p = L - d/2 and q = L + d/2. Derive equation: C f = L^2 - d^2/4L. Explain mathematically why no image is formed if the object and image are separated by a distance that is less than 4f. Solution A. given: 1/f = 1/p + 1/q which gives f = pq/(p+q) p =(L - d)/2 and q = (L+d)/2 f = [(L - d)/2]*[(L + d)/2]/[(L - d)/2 + (L + d)/2] f = (L^2 - d^2)/4L B. u = object distance, v = image distance s = separation of object and image, f = focal length of the lens. 1/u + 1/v = 1/f 1/u + 1/(s - u) = 1/f (s-u+u)/(u*(s-u)) = 1/f fs = su - u^2 u^2 = (u - f)s s = u^2/(u - f) By differentiating it ds/du = ((u - f)(2u) - u^2)/(u - f)^2 = (u^2 - 2uf)/(u - f)^2 = u(u - 2f)/(u - f)^2 ds / du = 0 when: u(u - 2f) = 0 u = 0 or, for a real image, u = 2f. When u < 2f, ds / du < 0. When u > 2f, ds / du > 0. u = 2f is therefore a minimum. By same method v = 2f is therefore a minimum image distance. which means total separation 4f..

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