An insect that is 4.90 mm long is placed 11.0 cm from a converging lens with a focal length of 12.0 cm.?

An insect that is 4.90 mm long is placed 11.0 cm from a converging lens with a focal length of 12.0 cm.


(a) What is the position of the image?


cm from the lens


(b) What is the size of the image?


mm


(c) Is the image upright or inverted?


inverted


upright





(d) Is the image real or virtual?


real


virtual





(e) What is the angular magnification if the lens is close to the eye? (Assume the user's near point to be 25 cm.)

An insect that is 4.90 mm long is placed 11.0 cm from a converging lens with a focal length of 12.0 cm.?
Let the object distance be p, image distance be q, and focus length be f. We have: 1/p + 1/q = 1/f





(a) What is the position of the image? (cm from the lens)


q = 1/(1/f-1/p) = 1/(1/12.0-1/11.0) = -132 (cm)





(b) What is the size of the image?


4.90*(-q/p) = 4.90*(132/11.0) = 58.8 (mm)





(c) Is the image upright or inverted?


upright





(d) Is the image real or virtual?


virtual





(e) What is the angular magnification if the lens is close to the eye? (Assume the user's near point to be 25 cm.)


25/12.0 = 2.08
Reply:maximum angular magnification = 1+ 25cm/f = 1+ 25cm/12cm = 3.08. Report It


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