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 S_{1 }S_{2} and object = D

focal length = *f*

Now, lens formula,

For a real image,

*u* → (–) ve

*v* → (+) ve

So,

Also,

*v* + *u* = D

=> *v* = D – *u*

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

Þ *u*D – *u*^{2} = *f*D

Þ *u*^{2} – D*u* + D*f* = 0

So,

**Case (i)** If D^{2} > 4D*f* i.e. __D > 4 f__ both roots

*u*

_{1}&

*u*

_{2}are real. That means convex lens give two positions

*u*

_{1}&

*u*

_{2}for which real images are formed on the screen.

**Case (ii)** when D^{2} = 4 D*f* Þ D < 4*f*, both roots are imaginary, so in this case no real image is formed.

**Case (iii) **when D^{2} 4D*f* Þ D = 4*f*, 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,

So, minimum distance between the object & screen for formation of real image to be D ³**4 f**.

Now,

When lens at point P_{1}, it forms a real inverted & enlarged image (A_{1}B) of object at AB.

P_{1}B = *u*_{1}, P_{1 }B_{1} = *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 P_{2} of lens is obtained for which real inverted but diminished image A_{2} B_{1} is formed.

Now, BP_{2 }= *v*, P_{2}B_{1} = *u*

Also,

P_{1}P_{2} = *d*

Then *v* – *u* = *d *…… (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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