Question

An object arrow 2 cm high is placed 20 cm from a converging lens of focal length of 10 cm. Find a) The location of the final image, b) the nature of image, c) the magnification and height of the
image. d)Draw a ray diagram of the image formation

Using the same lens, work out the results for an Object distance of 8 cm. Answer Parts (a) through (d) as in problem

Answer 1

(a) Given, Object hei height, h = 2cm 8th of the conve conviogingen gens, f= +10 cm object distance, do = -20cm focal length From the thin leas formula, *-*. 5 +5. to I f 1 20 2-1 20 2- - D et di = 20 cm The image from the is located at a distance 200m rens at opposite side of the object. b) The is real C is inverted image and inverted ! The magnification is- Linacotes that - image The height of the image hi = ml x h. = 1X2 = 2 cm di do 20 20 ) is

dy Ray diagram: Lens converging F Image object + 201m 2olm @ @) f= 10 cm do - 8cm from thin di lens formula, [ + do to - 4-5 - D 11 8 40 di = 40 cm distance 40 cm The ht a image is located the lens at the same side of the object. from virtual The image is I boy and upright - =S Saya (c) 8 magnification, The height of the image is, ki = mx ho = 5x2 = 10 cm

а) Ray diagram: lens converging F object image ( 8 см 4otm
Thanks