Question

Problem 47. An object is placed 6 cm from a converging lens with a 5-cm focal length Plane of lens 5cm Ray enters lens parallel o axis and passes through focal point on other side Ray passing through lens plane on axIS IS undeflected. (a) Use the thin-lens equation (Eq. 14.64) to calculate the image distance. (b) The magnification of the image is given by m - -v/iu (A negative magnification implies an inverted image.) What is the magnification for the image in part (a)? A value Im| > 1 implies a"magnified" image. This is how a slide projector works.

Answer 1

Thin lens formula is given by 1/v - 1/u = 1/f

where u is the distance of the image

v is the distance of the object

f is the focal length

Thin lens formula is given by 1/v - 1/u = 1/f where u is the distance of the image v is the distance of the object f is the focal length Given u = -6 cm, f = + 5 cm, V = ? this lens eq^n: 1/V - 1/U = 1/f implies 1/V - 1/(-6) = 1/(+ 5) implies V = + 30 cm Answer theta is the image distance is 30 cm magnification (m) v/u = 30 cm/ - 6cm = -5 Answer The magnification is negative So the image is inverted Also Vertical-bar m Vertical-bar = Vertical-bar - 5 Vertical-bar = 5 > 1 that is Vertical-bar m Vertical-bar > 1 so the image is magnified