The minimum distance between an object and it | vedclass

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(B) Let the distance between the object and the real image be $d$.Let the distance of the object from the lens be $u = -x$ (using sign convention).The

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$4 f$

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(B) Let the distance between the object and the real image be $d$.Let the distance of the object from the lens be $u = -x$ (using sign convention).Then the distance of the image from the lens is $v = d - x$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{d - x} - \frac{1}{-x} = \frac{1}{d - x} + \frac{1}{x}$.Simplifying the equation: $\frac{1}{f} = \frac{x + d - x}{x(d - x)} = \frac{d}{dx - x^2}$.Rearranging into a quadratic equation: $x^2 - dx + fd = 0$.For $x$ to be a real value,the discriminant must be greater than or equal to zero: $D = b^2 - 4ac \geq 0$.$(-d)^2 - 4(1)(fd) \geq 0$.$d^2 - 4fd \geq 0$.$d(d - 4f) \geq 0$.Since $d > 0$,we must have $d \geq 4f$.Therefore,the minimum distance between the object and its real image is $4f$.

The minimum distance between an object and its real image formed by a convex lens is

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