For this picture, is there a better way to determine object extra-focal distance

Question

For this picture, is there a better way to determine object extra-focal distance is positive other than what I did at the top with LM (it's given on another page that image is erect, which would mean -f/x is positive since -f/x = LM and LM s positive).

My confusion is that it's all in negative space. Why would you not do -0.667 - (-0.25) = -41.67?

tiuc A virtual image is formed 40 cm from a thin lens when a real object is placed 25 cm from the lens. (Questions 2, 3 and 4) 3. What is object extra-focal distance? a. +10.77 cm b. +40.00 cm +41.67 cm d·+66.67 cm e. None of the above since x is a negaative number 1.SO D .66-0.2s: 0.417 fi 0.663 046. b.as en oh nefur) For

Explanation / Answer

distance can't be negative.

only if we apply them in formula, then sign convention is used to determine the respective position of the object and images.

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