Question

(a) The relation between the object distance S, the image distance S' and the focal length f of a lens producing a real image is given as 1 1 1 S S This is called the Gaussian form of the thin lens formula. Another form of the formula, the Newtonian form, is obtained by considering the distance x, from the object to the first focal point and the distance x; from the second focal point to the image. Show that x.xx: - f2 is the Newtonian form of the thin lens formula. (b) A luminous object and a screen are placed at a fixed distance D apart. A converging lens of focal length f (f <D/4), placed between the object and the screen will form real image on the screen for two different lens positions that are separated by a distance x. (i) Show that the focal length of the lens is given by f - (D2 - x2) 4D (ii) If the image sizes for the two positions are le and Ir, then the size of the object is 1. - VTEŘIR

Answer 1

For part b. The solution becomes much easier if we use the basic knowledge of the principle of reversibility of light which states that light will follow exactly the same path if its direction of travel is reversed.

In other words if we place the object at the place of screen( image) then the image will form at the place of object. Now proceed to see th detailed solution in the image attached.
IL -P ENS 2 a) a to . TI f IMAGE ي ع ب OBJEG f VE TE WE HAVE, I V u = 1 f es - 55 $ in EQUATION O put (f+20) = p all - (f+x1) = 89 - + = | 1 . 2 Xot f 1 xi tf = - a (20+) (Xi+f) = f( Xot X; +2f) Xoxi + Xef + fx; + f²= foto + fx; tzf- 20 x; =f?/ HENCE PROVED, x w SCREEN LENS LENS OBJECT POSITION POSITION POSITION USING THE PRINCIPLE OF REVERSIBILITY OF UGHT 0 D = utv OC = V-U - - SOLVING JUD -2 O&@ WE GET v= D+X tin 2

þ using USING LEN'S FORMULAE I CORRECTION dD II V u. LEA 2 - Z dtx (d-s() f J af [d-x + d + x ] = d. -4fol = x² F - D2-x? 7 HENCË PROVED ! FOR POSITION 0 S ur FOR POSITION ② simi - - M2 = 4 og ALSO MI = IE ALSO, M2 = IR ""E from EQUATIONS ® & © ( amim -viy t mim2 = 1 saulga " > Io = √ EXIR