Question

f- +20 cm ------ So Yo 30 20 L. 43 30 10 70 60 3 So 80 CH 70 cm A real object 15 cm tall is 50 cm to the lest of a leus of focal length +20 en Calculate and determine (4) image distance di Auslal di (6) nature of image Ans(circle-owe real virtual (c) orientation of image Ausly civile owe: erett inverted! (d) image height Ansleff his (e) [2 points extra creded] On the picture above, show a ray (3 standerd rays) diagram use it to locate the image. <to lograded right or winy; it has to be perfect. ② No eguations needed here but not forbidden either reya 5 15 10 35 30 20 20 30 40 го 70 40 50 40 raya ray o Rays A, B, and, emanating from the tip of a real object at 25cm, pass through a lens of as-yet-sukupum shape and art beat as shown (a) what kind of image is the leas forming ? auscul, civede one real virtual (6) On the diagram, sketch the image is), at the correct location (a) whether right side up for upside downt (c) What approximate numerical value of the image distance d; should be used in the thin lens equation? Au (c) di = seletas circle (a) What kind of lews is it? Converging + се Sign diversing I one Pos F.

Answer 1

7. (b) As the The lens equation is the koe sotto U= 33.33 cm (di) image distance is tve, magnification, 쁩 As the magnifications in is-ve, image is inverted. image height, hi = /mlx ho z 0.667x15 ² 10 cm the image is real. ma -33.33 50 z-0.667 u (d) (e) Im oboj 8. (b) day A Image lage B la) The image is virtuel. le) Approrimete image distance is, di=-67 cm Converging (conver) lens. (d) The lem is a Cone