Find the image distance - optical instruments

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Homework Help Overview

The problem involves determining the image distance for an object placed in front of a diverging lens, given its focal length. The context is within the subject area of optics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the lens formula but questions their calculation after receiving a different answer. Some participants suggest checking for calculation errors and emphasize the importance of rearranging equations for clarity.

Discussion Status

The discussion is ongoing, with participants exploring the calculations involved and addressing potential errors. Guidance has been offered regarding the rearrangement of equations to better understand expected outcomes.

Contextual Notes

There is a noted discrepancy between the original poster's calculated image distance and the answer provided to them, prompting further inquiry into the assumptions and calculations made.

ally1h
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Homework Statement


An object is placed at 30 cm in front of a diverging lens with a focal length of 10 cm. What is the image distance?


Homework Equations


1/f = 1/do + 1/di



The Attempt at a Solution


I thought this was a straight-forward question:
1/di = 1/-f - 1/do
1/di = (1/-10 cm) - (1/30 cm)
1/di = -0.066 cm
di = 1/-0.066 cm = -15 cm

BUT- the answer provided to me states that the answer is -7.5cm. So what is it that I'm missing here?
 
Physics news on Phys.org
Slight calculator typo
(1/-10) - (1/30) = - (1/10 + 1/30) = -0.133
 
Well now... don't I feel silly?

Thank you.
 
It's very easy to get those sort of sums wrong,
It's worth rearranging them so that you know if the answer you expect is +ve or -ve
And in a form where you can estimate the magnitude, eg (1/10+1/30) is obviously going to be a bit bigger than 1/10
 

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