Converging lens solve for all variables given f, Hi, and Ho

[Samr28]

Homework Statement

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What is directly given in the problem:
Converging lens with f=1.43m
Hi = 2m

Homework Equations

Hi/Ho = di/do
M = Hi/Ho = di/do
1/f = 1/do + 1/di

The Attempt at a Solution

Ho = 2m * 20 = 40m
Not sure how I can get do from this. All of the equations seem to need di.

 

[Daniel Gallimore]

Homework Statement

[/B]
View attachment 200090

What is directly given in the problem:
Converging lens with f=1.43m
Hi = 2m

Homework Equations

Hi/Ho = di/do
M = Hi/Ho = di/do
1/f = 1/do + 1/di

The Attempt at a Solution

Ho = 2m * 20 = 40m
Not sure how I can get do from this. All of the equations seem to need di.

You can get the object height from the magnification. Since the image is 20 times smaller than the object, M=1/20. We also know that H_i=2 \, m.

To get the distance to the coaster, multiply \frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i} through by d_o and substitute 1/M=20 for d_o/d_i. You know the focal length, so it shouldn't be difficult to solve for d_o from there.

 

[andrevdh]

the image is 20x smaller than the object so

h_i = 1/20 h_o

giving us

h_i/h_o = 1/20

but h_i/h_o = d_i/d_o

so we also have that

d_i/d_o = 1/20

so that we have that the object distance in terms of the image distance is

d_o = ... ?

 

[Daniel Gallimore]

the image is 20x smaller than the object so

h_i = 1/20 h_o

giving us

h_i/h_o = 1/20

but h_i/h_o = d_i/d_o

so we also have that

d_i/d_o = 1/20

so that we have that the object distance in terms of the image distance is

d_o = ... ?

Multiply the equation \frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i} through by d_o. This gives you the equation \frac{d_o}{f}=\frac{d_o}{d_o}+\frac{d_o}{d_i} d_o/d_o=1 and d_o/d_i=1/M Solve for the remaining d_o.