Question

What is the minimum distance between an object and its real image formed by a convex lens?​

Answer 1

Answer:

according to distinct distance of vision it is 25cm

Answer 2

SOLUTION:-

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For a convex lens we have

$\boxed{ \: \: \sf - \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \: \: \: \: \: \: \: \: \: ....(1) }$

Now suppose that an object is placed at a distance x from the other convex lens and its real image is formed on the other side of the lens at a distance u from it.

Now, u=-x ;v=+v and f=+f

Therefore the first equation becomes

$\sf:\implies - \dfrac{ \: \: 1 }{ - x} + \dfrac{1}{ + v} = \dfrac{1}{ + f}$

$:\implies \sf \: \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{x}$

$:\implies \sf \: \dfrac{1}{v} = \dfrac{x - f}{fx}$

$:\implies \sf \: v = \dfrac{fx}{x - f} \: \: \: \: ....(2)$

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If D is the distance between the object and a image then

⇒v=D-x ———(3)

From the equations (2) and (3) , we have

$\sf : \implies \: D - x = \dfrac{fx}{x - f}$

$\sf : \implies \: D = x +\dfrac{fx}{x - f} \: \: \: \: ....(4)$

Differentiating both sides of the equation (4) with respect to x, we have

$\sf: \implies\dfrac{dD}{dx} = 1 + \dfrac{f}{x - f} + fx( - 1)(x - f)^{ - 2}$

$\sf: \implies\dfrac{dD}{dx} = 1 +\dfrac{f}{x - f} - \dfrac{fx}{(x - f) ^{2} }$

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For D to be minimum,

$\sf: \implies\dfrac{dD}{dx} = 0$

$\sf :\implies1 + \dfrac{f}{x - f} - \dfrac{fx}{(x - f) ^{2} } = 0$

$\sf : \implies \sf \: \dfrac{x}{x - f} = \dfrac{fx}{(x - f)^{2} }$

$\sf : \implies\dfrac{f}{x - f} = 1$

$: \implies \sf \: x = 2f$

In the equation (4) , Substituting x=2f , we have

$\sf : \longmapsto \: \: D = 2f + \dfrac{f \times 2f}{2f - f} = 2f + 2f = \bf \: 4f$

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