Question

An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the image. Is the image upright or inverted? Is it real or virtual?

Answer 1

Answer:

Inverted

Real

Explanation:

u = Object distance =  30 cm

v = Image distance

f = Focal length = 10 cm

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm$

As, the image distance is positive the image is real and forms on the other side of the lens

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-15}{30}\\\Rightarrow m=-0.5$

As, the magnification is negative the image is inverted