Thin Lens And Lens Maker Equations Video Lessons

Concept

# Problem: Rank the images on the basis of the magnitude of their magnification, from greatest to smallest.Rank the images on the basis of their size, assuming the object size is the same. (in cm)1. d=15; f=202. d=15; f=53. d=5;   f=204. d=20; f=105. d=10; f=206. d=10; f=5

###### FREE Expert Solution

Lens maker equation:

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

Magnification,

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

From the lens maker equation:

$\begin{array}{rcl}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{o}}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{f}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{d}}\mathbf{=}\frac{\mathbf{d}\mathbf{-}\mathbf{f}}{\mathbf{fd}}\\ \mathbf{m}& \mathbf{=}\mathbf{-}& \frac{\mathbf{d}}{{d}_{o}}\\ & \mathbf{=}& \mathbf{-}\mathbf{d}\mathbf{\left(}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{o}}}\mathbf{\right)}\\ & \mathbf{=}& \frac{\mathbf{-}\overline{)\mathbf{d}}\mathbf{\left(}\mathbf{d}\mathbf{-}\mathbf{f}\mathbf{\right)}}{\mathbf{f}\overline{)\mathbf{d}}}\\ & \mathbf{=}& \frac{\mathbf{f}\mathbf{-}\mathbf{d}}{\mathbf{f}}\end{array}$

The sizes of the images are determined by the magnification.

###### Problem Details

Rank the images on the basis of the magnitude of their magnification, from greatest to smallest.

Rank the images on the basis of their size, assuming the object size is the same. (in cm)

1. d=15; f=20

2. d=15; f=5

3. d=5;   f=20

4. d=20; f=10

5. d=10; f=20

6. d=10; f=5