DISPLACEMENT METHOD TO DETERMINE THE FOCAL LENGTH OF A CONVEX LENS

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DISPLACEMENT METHOD TO DETERMINE THE FOCAL LENGTH OF A CONVEX LENS

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If the distance D of an object between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens between the object and the screen at which a sharp image of the object is formed on the screen. This method is called displacement method.

Let U = distance of an object from a convex lens.

F = focal length.

V = D – U = the distance of image from the lens.

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From lens formula

\begin{array}{l} \frac{1}{{\rm{v}}}\, - \,\frac{1}{u}\, = \,\frac{1}{f}\\ \frac{1}{{D - u}}\, - \,\left( {\frac{1}{{ - u}}} \right)\, = \,\frac{1}{f}\\ \frac{1}{{D - u}}\, + \,\frac{1}{u}\, = \,\frac{1}{f} \end{array}

On solving

\begin{array}{l} {u^2}\, - \,Du\, + \,Df\, = \,0\\ u\, = \,\frac{{D\, \pm \,\sqrt {{D^2}\, - \,4\,fD} }}{2} \end{array}

Now, there are following possibilities:

(a) If D < 4 f, then u is imaginary.

So, physically no position of the lens is possible.

(b) If D = 4 f, thenu\, = \,\frac{D}{2}\, = \,2f. So only one position is possible. The minimum distance between an object and its real image in case of a convex lens is 4 f.

(c) If D > 4 f, there are two positions of the lenses at distances{u_1}\, = \,\frac{{D\, - \sqrt {{D^2}\, - \,4\,fD} }}{2} and {u_2}\, = \,\frac{{D\, + \sqrt {{D^2}\, - \,4\,fD} }}{2}

Since v = D –u. Therefore

{{\rm{v}}_1}\, = \,\frac{{D\, + \sqrt {{D^2}\, - \,4\,fD} }}{2}

{{\rm{v}}_2}\, = \,\frac{{D\, - \sqrt {{D^2}\, - \,4\,fD} }}{2}

Let the lens be displaced through x,

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\begin{array}{l} x\, = \,{{\rm{v}}_1}\, - \,{u_1}\, = \,\sqrt {{D^2} - \,4\,fD} \\ {x^2}\, = \,{D^2}\, - \,4fD\\ f\, = \,\frac{{{D^2}\, - \,{x^2}}}{{4D}} \end{array}

Let {m_1}and {m_2}be the magnification for the first position of lens {L_1}and second position of the lens{L_2}.