DISPLACEMENT OF A LENS WHEN OBJECT AND SCREEN BOTH ARE FIXED.

This method is to know as DISPLACEMENT METHOD and is used in the laboratory to determine the focal length of the convex lens. In this case, if the object is at a distance. u from the lens, the distance of the image from the lens v=D-u so, from the lens- mirror formula

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

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

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

\frac{\not{u}+D-\not{u}}{\left ( u \right )\left ( D-u \right )}=\frac{1}{f}

Df=uD-u^{2}

u^{2}-uD+Df=0\;\;\;\;\;\;\;\;\;\;\;\left ( 1 \right )

u=\frac{\left ( -D \right )\pm \sqrt{\left ( -D \right )^{2}-4Df}}{2\times 1}

u=\frac{D\pm \sqrt{D^{2}-4Df}}{2}

u=\frac{D\pm \sqrt{D\left ( D-4f \right )}}{2}

Now, there are different conditions.

(a)  if D< 4f;\;\;uwill be imaginary, so physically no position of the lens is possible

(b)  if D= 4f; in this situation u=\frac{D}{2}=\frac{4f}{2}=2f so, only one position is possible and in this situation v=D-u=2f.

(c)  if D> 4f; in this case both roots of equation (1) will be real.

u_{1}=\frac{1}{2}\left [ D-\sqrt{D\left ( D-4f \right )} \right ] and u_{2}=\frac{1}{2}\left [ D+\sqrt{D\left ( D-4f \right )} \right ]
• so, if d> 4f, there are two positions of lens at distance u_{1} and u_{2} from the object for which real image is formed on the screen.

if the distances bet ween two positions of the lens is d

\therefore \;\;d=u_{2}-u_{1}=\sqrt{D\left ( D-4f \right )}

\therefore \;\;d^{2}=D\left ( D-4f \right )

\Rightarrow \;\;d^{2}=D^{2}-4Df

\Rightarrow \;\;4Df=D^{2}-d^{2}

\Rightarrow \;\;f=\frac{D^{2}-d^{2}}{4D}

• the image distances corresponding to two positions of the lens will be
v_{1}=D-u_{1}

\therefore \;\;v_{1}=D-\frac{1}{2}\left [ D-\sqrt{D\left ( D-4f \right )} \right ]

\therefore \;\;v_{1}=\frac{1}{2}\left [ D+\sqrt{D\left ( D-4f \right )} \right ]

\;\;\;\;\;v_{1}=u_{2}

similarly,

v_{2}=D-u_{2}

\therefore \;\;v_{2}=D-\frac{1}{2}\left [ D+\sqrt{D^{2}-4Df} \right ]

\therefore \;\;v_{2}=\frac{1}{2}\left [ D-\sqrt{D^{2}-4Df} \right ]

we see that,

\;\;\;\;\;v_{2}=u_{1}

i.e; for two positions of the lens object and image distances are interchangeable.

Again, we have, d=\sqrt{D(D-4f)} \therefore \;\;\;u_{_{1}}=\frac{1}{2}\left [ D-d \right ]

\therefore \;\;\;u_{_{1}}=v_{2}

&\;\;\;u_{_{2}}=\frac{1}{2}\left [ D+d \right ]

\;\;\;u_{_{2}}=v_{1}

magnification in the first position of the lens

\;\;\;m_{_{1}}=\frac{v_{1}}{u_{1}}

\;\;\;m_{_{1}}=\frac{ \left ( h_{im} \right )_{_{1}}}{h_{\circ}}

Similarly, magnification in the 2nd position of the lens

\;\;\;m_{_{2}}=\frac{v_{2}}{u_{2}}

\;\;\;m_{_{2}}=\frac{ \left ( h_{im} \right )_{_{2}}}{h_{\circ}}

Now,  (a) \;\;\;m_{1}m_{2}=\frac{v_{1}}{u_{_{1}}}\times\frac{v_{2}}{u_{2}}

\;\;\;m_{1}m_{2}=\frac{\left ( h_{im} \right )_{1}}{\left ( h_{_{\circ}} \right )}\times\frac{\left ( h_{im} \right )_{2}}{\left ( h_{_{\circ}} \right )}

\;\;\;m_{1}m_{2}=1

\Rightarrow \;\;\;m_{1}m_{2}=\frac{\left ( h_{im} \right )_{1}\left ( h_{im} \right )_{2}}{\left ( h_{_{\circ}} \right )}

\therefore \;\;\;h_{\circ}= \sqrt{\left ( h_{im} \right )_{1}\left ( h_{im} \right )_{2}}

(b)  \frac{m_{1}}{m_{2}}=\frac{\frac{v_{1}}{u_{1}}}{\frac{v^{_{2}}}{u_{2}}}

\therefore \;\;\frac{m_{1}}{m_{2}}=\frac{u_{2}}{v_{1}}\times \frac{v_{2}}{u_{1}}

\therefore \;\;\frac{m_{1}}{m_{2}}=\frac{u_{2}^{2}}{v_{1}^{2}}

\therefore \;\;\frac{m_{1}}{m_{2}}=\left ( \frac{D+d}{D-d} \right )^{2}

(c) \;\;\;m_{1}-m_{2}=\frac{V_{1}}{U_{1}}-\frac{V_{2}}{U_{2}}

\;\;\;\;\;\;m_{1}-m_{2}=\frac{U_{2}}{U_{1}}-\frac{U_{1}}{U_{2}}

\;\;\;\;\;\;m_{1}-m_{2}=\frac{U^{2}_{2}-U^{2}_{1}}{U_{1}U_{2}}

\therefore \;\;m_{1}-m_{2}=\frac{\left ( \frac{D+d}{2} \right )^{2}-\left ( \frac{D-d}{2} \right )^{2}}{\left ( \frac{D-d}{2} \right )\left ( \frac{D+d}{2} \right )}

\therefore \;\;m_{1}-m_{2}=\frac{4Dd}{D^{2}-d^{2}}

Also, \;\;f=\frac{D^{2}-d^{2}}{4D}

\;\Rightarrow{D^{2}-d^{2}}=4fD.

\therefore \;\;m_{1}-m_{2}=\frac{\not{4}\not{D}d}{\not{4}f\not{D}}

\therefore\;\;m_{1}-m_{2}=\frac{d}{f}

\therefore\;\;f=\frac{d}{m_{1}-m_{2}}

Q.    A convex lens forms an image of an object on a screen. The height of the image in 9 cm. The lens is now displaced until an image is again obtained on the screen. the height of this image is 4 cm . The distance between the object and the screen is 90 cm.

(A) The distance between the two poritns of the lens is 30 cm.

(B) The distance of the object from the lens in its first position is 36 cm.

(C) The height of the object is 6cm.

(D) The focal length of the lens is 21.6 cm.

Ans.  As, we know,

\sqrt{\left ( him \right )_{1}\left ( him \right )_{2}}=h_{\circ}

\left ( him \right )_{1}=9 \;cm ,\;\;(him_{2})= 4\; cm

\therefore\;\;m_{1}=\frac{\left ( him \right )_{1}}{h_{\circ}}

\Rightarrow \;\; m_{1}=\frac{9}{6}

\Rightarrow \;\; m_{1}=\frac{3}{2}

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