Question

12) Two lenses are placed 13 cm apart. The first lens is a converging lens with focal length 4cm. The second lens is a diverging lens with focal length 9cm. If an object is placed at a distance of 13 cm from the first lens, A) What is the position of the image? (6 points) B) What is the image type (virtual or real)? (2 points) C) What is the magnification? (2 points)

Answer 1

find image distance by using only lens -1 1 - | fi do, diz .) 4 13 dig Cm Cancro diz = 5.70 (real image) Magnification, -dis mia doa haitos 0.44 (inverted image) Now, image position by lens-2 do₂ = 13-dia 13-5.72 do = m, ) 7.22 cm + fz do, diz

= -9 7.22 diz diz =- cm ( image is vitual and same side of the object) Magnification - diz m₂ do - (-4) 7.22 m. - 0.55 A) 4 cm Position of the image is to the left side of diverging lens go is virtual . B) mage c) Magnification. m = mixm - 40.44) * 0.55 m -0.24 f2 fi fz fi 6