Question

QUESTION 4 Two converging lenses having focal lengths of f1 = 10.5 cm and f2 = 18.0 cm are placed a distance d = 52.5 cm apart as shown in the figure below. The image due to lig passing through both lenses is to be located between the lenses at the position x = 30.5 cm indicated. Object Final image position -d a) At what value of p (in cm) should the object be positioned to the left of the first lens? QUESTION 5 b) What is the magnification of the final image?

Answer 1

- Solo Givens focal length fi = 10.5cm focal length fa= 18.00m distance d = 52.5cm position a = 30.5cm Ray diagram: image object Object A HI T X = 30.5cm -S * let H> Height of thest image hy Height of final image he height of object qa distance of first image from dens fi s> distance of first image from dens fa s distance of final Image from lens fa (0) Now we use lents formula for first hoge line su t fp= 18.0 cm :-sed- = 52.5 - 30.5 -S'= 22 → s'= -22 CS Scaniad with Scan with the to CamScanner

Śs hat to s= 22.2.+18 22X18 the S = 9.9 cm the position of the efforst Image with respect to birst fange dens a 2= dos 2 = 52.5-9.9 q = 42.6 cm Now use dens formula for final loage lens $# $= 1 = tos 1226 TP = 8.42 cm ] Any. (b) Position of object to the left of the first dens po ſp= 8.42cm Any magnification of the final finage - -0 -for frese image H = 1/2 @ for final image Scanned with CamScanner

from feqleetion o M= equation ® = Ma s. 2 M= - 11.24] Am sign indicates that final image is invested. Negative if you are satisfy with with like if not then are happy to help you Scanned with CamScanner us then please appricate us tell us in comment box, we again. Thank you.