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  • In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen.

In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

 

Answers (1)

 This method is known as the "Displacement method" is also used to find the confocal length of the length in laboratory

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

If we take a convex lens L place between an object O and a screen S. The distance between the object and the screen would be D and the position of the object and screen are held fix. Let the distance of position 1 from the object be x_{1}. Then the distance of the screen from the lens is D-x_{1}.

Therefore, u=-x_{1} and v=+\left ( D-x_{1} \right ).

Placing it in the lens formula,

\frac{1}{D-x_{1}}-\frac{1}{\left ( -x_{1} \right )}=\frac{1}{f}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(i)

At position II, let the distance of the lens from the screen be x_{2}. Then the distance of the screen from the lens is D-x_{2}

Therefore, u=-x_{2} and v=+\left ( D-x_{2} \right ).

Placing it in the lens formula

\frac{1}{D-x_{2}}-\frac{1}{\left ( -x_{2} \right )}=\frac{1}{f}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(ii)

Comparing Eqs. (i) and (ii), we realize that there are only two solutions:

1.x_{1}=x_{2}; \text {or}

2.D-x_{1}=x_{2}; \text {and}\; D-x_{2}=x_{1}

The first solution is trivial. Therefore, if the first position of the lens, for a sharp image, is x_{1} fro the object, the second position is at D-x_{1} from the object.

Let the distance between the two positions I and II be d.

From the diagram, it is clear that

D=x_{1}+x_{2}\; \text {and}\; d=x_{2}-x_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(iii)

On solving, two equations in (iii) we have

x_{1}=\frac{D-d}{2} and D-x_{1}=\frac{D+d}{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(iv)

Substituting Eq. (iv) in Eq. (i), we get

\frac{1}{\left ( \frac{D+d}{2} \right )}-\frac{1}{\left ( -\frac{D-d}{2} \right )}=\frac{1}{f}\Rightarrow f=\frac{D^{2}-d^{2}}{4D}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(v)

or d=\sqrt{D^{2}-4Df}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(vi)

If u=\frac{D}{2}+\frac{d}{2}, then the image is at v=\frac{D}{2}-\frac{d}{2}.

\therefore The magnification m_{1}=\frac{D-d}{D+d}

If u=\frac{D-d}{2}, then v=\frac{D+d}{2},

\therefore The magnification m_{2}=\frac{D+d}{D-d}

Thus, \frac{m^{2}}{m_{1}}=\left ( \frac{D+d}{D-d} \right )^{2}

This is the required expression of magnification.

Posted by

infoexpert23

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