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Question

Physics Question on Ray optics and optical instruments

In a convex lens of focal length FF, the minimum distance between an object and its real image must be

A

3F

B

4F

C

32\frac{3}{2}

D

2F

Answer

4F

Explanation

Solution

Let L is the distance between a real object and its real image formed by a convex lens, then as L=(u+v)L = \left(|u|+|v|\right) =(uv)2+2uv...(i)\,=\left(\sqrt{u}-\sqrt{v} \right)^2+2\sqrt{uv}\quad\quad\quad\quad ...\left(i\right) L will be minimum, when (uv)2=0\left(\sqrt{u}-\sqrt{v}\right)^2=0 i.e.,u=vi.e., u=v Putting, u=uu = -u and v=+uv = +u in lensformula, 1v1u=1F\frac{1}{v}-\frac{1}{u}=\frac{1}{F} 1u1u=1F\frac{1}{u}-\frac{1}{-u}=\frac{1}{F} u=2Fu = 2F (L)min=22F×2F=4F(Using(i))\therefore\, \left(L\right)_{min}= 2\sqrt{2F\times2F}=4F \quad\left(Using \left(i\right)\right)