Calculating Effective Focal Length | Mirror Equation Help - 19mm Answer

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The discussion centers on calculating the effective focal length of the eye's lens using the mirror equation. The user initially struggles with unit conversion, mistakenly using 40 cm instead of 400 mm for the object distance. Clarification is provided that the distance from the retina to the lens (20 mm) is the image distance, while the object distance is indeed 400 mm. The correct application of the formula 1/f = 1/do + 1/di leads to the answer of 19 mm, as stated in the book. The conversation highlights the challenges faced by students in understanding physics concepts, particularly in a difficult learning environment.
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Mirror equation help! please urgent!

Homework Statement



The distance from the lens of someone's eye to the retina is 20.0 mm. if the image of a book held 40.0 cm is in front of the eye in sharp focus, what is the effective focal length of the lense.
the back of the book says the answer is 19 mm.

thanks for your help everyone!

Homework Equations


1/f = 1/ do + 1/ di



The Attempt at a Solution

 
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radiofruit said:

Homework Statement



The distance from the lens of someone's eye to the retina is 20.0 mm. if the image of a book held 40.0 cm is in front of the eye in sharp focus, what is the effective focal length of the lense.
the back of the book says the answer is 19 mm.

thanks for your help everyone!

Homework Equations


1/f = 1/ do + 1/ di



The Attempt at a Solution

Just plug in the values.

Don't forget that 40 cm = 400 mm
 


Gahh! that's why it didn't work.

I forgot to convert Units.

Ahhh I'm a silly physics 11 student.

Thanks for your help!

Hmm does the distance from the retina to the lense (20 mm ) count as the Do ?
 


radiofruit said:
Gahh! that's why it didn't work.

I forgot to convert Units.

Ahhh I'm a silly physics 11 student.

Thanks for your help!

Hmm does the distance from the retina to the lense (20 mm ) count as the Do ?

You are welcome:smile:

No, 20 mm is Di, the distance to the image (which is on the retina). 400 mm is the object distance
 


Thanks!

So glad I join this message board.

My physics 11 teacher is horrendous, our class average is in the 50's.

I'm managing a 'B', but it's been difficult.
 


radiofruit said:
Thanks!

So glad I join this message board.

My physics 11 teacher is horrendous, our class average is in the 50's.

I'm managing a 'B', but it's been difficult.

You are welcome. Sorry to hear about your situation! I wish you the best luck!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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