Derive the formula :f = D(square) - d (square) / 4d where D is the distance between the object and the screen and d is ditance between two locations of convex lens and f is the focal length.

Condition for a real image formed by convex lens when object & screen are fixed.



Distance between screen S1 S2 and object = D

focal length = f


Now, lens formula,


For a real image,

u → (–) ve

v → (+) ve





v + u = D

=> v = D – u


Plug-in the value of v in equation (i)


Þ uD – u2 = fD

Þ u2 – Du + Df = 0





Case (i) If D2 > 4Df i.e. D > 4f  both roots u1 & u2 are real. That means convex lens give two positions u1 & u2 for which real images are formed on the screen.


Case (ii) when D2 = 4 Df Þ D < 4f, both roots are imaginary, so in this case no real image is formed.


Case (iii) when D2 4Df Þ D = 4f, two roots are real & equal.

So, in this case only a real image will be formed on the screen for only one position of the lens.


So, minimum distance between the object & screen for formation of real image to be D ³4f.





When lens at point P1, it forms a real inverted & enlarged image (A1B) of object at AB.

P1B = u1, P1 B1 = v

Then, u + v = D  ……(ii)


Lens formula,


For real image,

u → (–)ve

v → (+)ve



Here, u & v are symmetrical, so they may be interchanged.


Now, if we displace the lens towards screen PQ, then one more position P2 of lens is obtained for which real inverted but diminished image A2 B1 is formed.


Now, BP2 = v, P2B1 = u



P1P2 = d

Then vu = …… (iii)


From, (i) & (ii)


Using sign convention

u → (–)ve


v → (+)ve



Substituting these values in lens formula and solving then you get

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