Show that the least possible distance between an object and its real image in aconvex lens is 4f, where f is the focal length of the lens
Question
Show that the least possible distance between an object and its real image in a convex lens is 4f, where f is the focal length of the lens.
Solution
Sure, here is the step by step solution:
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The lens formula for a convex lens is given by 1/v - 1/u = 1/f, where v is the image distance, u is the object distance, and f is the focal length of the lens.
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We are looking for the minimum value of the distance between the object and its real image, which is |v - u|.
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To find the minimum value of a function, we can take its derivative and set it equal to zero. However, in this case, it's easier to complete the square.
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Rewrite the lens formula as 1/v = 1/f + 1/u.
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Then, rewrite it again as v = 1/(1/f + 1/u).
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Now, express u in terms of v and f: u = fv/(v - f).
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Substitute u back into the expression for |v - u| to get |v - fv/(v - f)|.
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Simplify this to get |(v^2 - 2fv + f^2)/(v - f)| = |(v - f)^2/(v - f)| = |v - f|.
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The minimum value of |v - u| = |v - f| occurs when v = 2f.
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Substitute v = 2f back into the lens formula to get 1/2f - 1/u = 1/f, which simplifies to u = 2f.
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Therefore, the minimum distance between the object and its real image is |2f - 2f| = 4f.
So, the least possible distance between an object and its real image in a convex lens is 4f, where f is the focal length of the lens.
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