Question

3. (a) An object is placed 15 cm in front of a convex lens of focal length 10 cm. 10 cm past the first lens is a concave lens of focal length 10 cm. 50 cm past the second lens is a convex lens of focal length 20 cm. With respect to the initial object, where would be the final image and what is its magnification? (Iz F = do di 10&M 50 cm Tot is to s to s Bo-30 di = 30cm fy=10cm +3 = 20 cm ③ / 2 = 1 + 1 15cm (+) (t = 10com do -20 07- lett 20 -70 di o di = - 20cm m=di = -1-20) = -1 id; = 28cm M=(-2) (-1) (-2) € - 35 +=žo di 20 7o - že

Where would be the final image and what is its magnification? Please check my work.

Answer 1

Image formation by Iet lene Ling sign convention. Object distanu, 4,5-151m focal length q convex lang, f = 100m Sinu t ti = t3 t o 1 = (tot). ci V, - 300m → Image formation by and lens Objet dus tance, U2 = (80-10) cm = 200m focal length, fx = -10am einu, t tot 12 - 16-to + 2 = (toto) 5 V23-2017 → Image formation by Brd ons Objet die tance, 43 = -(20+50)\ =-701m so cal length, fa= 200m to-toets tes - to > to hoto) shoes => y = 28 cm Thus with respect to initial object final image distance is d=(15+10+50+28) om 5103 cm d=103 em - Magnification, M= V x x 2 a 20 x 20 x 25 26 Ma-008