20 A diverging lens (f = -10 cm) is located 20.0 cm to the left of a converging

Question

20 A diverging lens (f = -10 cm) is located 20.0 cm to the left of a converging lens (f = 32.0 cm). A 3.00 cm tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. cm (b) What is the height of the final image (including proper algebraic sign)? cm

Explanation / Answer

................********................… ..|..............*.....*...............… --|---11cm..--*...*------20 cm.--*-----*------> x (+ value) .................*.....*...............… ................********...............… Now use the formula 1/di + 1/do = 1/f where di is the distance from the lens to the image, do is the distance from the object to the lens, and f is the focal distance of the lens object is at focal point of lense means do = 11, and f = -11, so we get 1/di + 1/11 = -1/11 1/di = -2/11 ; or di = -11/2 = - 5.5 cm This means that the image formed by diverging lens with a real object will be virtual image and would be same side -5.5 cm from the centre. The magnification M of a lens is the ratio of the image size or height to the object height. A negative magnification signifies an inverted image. By this convention, a positive height indicates the arrow is erect and a negative height means the arrow is pointing down (valid for both object and image). M = hi / ho = -(di / do) hi / 3 = - (-5.5 / 11) = hi = 1.5 cm positive means pointing upwards since first lense is 20 cm away from second converging lense so total distance (di) for converging lense would be 25.5cm and we know f = 32 cm, so again 1/di + 1/do = 1/f 1/di + 1/25.5 = 1/32 1/di = 1/32 - 1/25.5 or di = -125.53 cm (from the center of converging lense) --------------Answer since distance is negative means left side of the lense or same side where object was Image will be virtual and magnifiaction would be hi / ho = -(di / do) hi / 1.5 = - (-125.53 / 25.5) or hi = 7.38 cm +ve means upward, (vertical), virtual image same side, at 125.53 cm

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