Thin Lens And Lens Maker Equations Video Lessons

Concept

# Problem: An object is placed 10 cm to the left of a converging lens that has a focal length of 20 cm. Describe what the resulting image will look like (i.e) image distance, magnification , upright or inverted images, real or virtual images?)

###### FREE Expert Solution

Lens equation:

$\overline{)\frac{\mathbf{1}}{{\mathbit{s}}_{\mathbit{o}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbit{s}}_{\mathbit{i}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbit{f}}}$

Magnification:

$\overline{){\mathbf{m}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{s}}_{\mathbf{i}}}{{\mathbf{s}}_{\mathbf{o}}}}$

From the Lens equation:

1/si = 1/f - 1/so

###### Problem Details

An object is placed 10 cm to the left of a converging lens that has a focal length of 20 cm. Describe what the resulting image will look like (i.e) image distance, magnification , upright or inverted images, real or virtual images?)