q) What is the minimum distance between and object and it's real image for a convex lens of focal length 'f'.

Since the image is real, minimize subject to .

its certainly 4f ..

ie  when object is  at centre of curvature

DISPLACEMENT METHOD:



Let us consider a real object and a screen fixed at a distance D apart. A convex lens is placed between them.

 

Let x be the distance of the object from the lens when its real image is obtained on the screen.

we have, u = - x

             v = + (D - x)

             f = + f

Using lens formula, we have

     

or 

=> x² - Dx + Df = 0

or  x =

Cases:-



(i) If D < 4f, for no position of lens, image formation is possible.



(ii) If D = 4f, then in this case x = = 2f. So only one position of the lens will be possible.



(iii) If D > 4f, in this situation, two positions of lens will give same image position. The positions are x₁ =     x₂ =

x₂ - x₁ = L =

ITS DEFINATELY NOT  "4f". AS  4F IS THE DISTANCE OF THE OBJECT TO ONLY GET A VERY SHARP IMAGE.

FOR REAL IMAGE 1/U+1/V=1/F

FOR VIRTUAL IMAGE 1/U-1/V=1/F

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