An insect that is 4.20 mm long is placed 9.2 cm from a converging lens with a fo

Question

An insect that is 4.20 mm long is placed 9.2 cm from a converging lens with a focal length of 12.0 cm. (a) What is the position of the image? distance Correct: Your answer is correct. 39.4 cm from the lens (q=-39.4) location Correct: On same side of object. Your answer is correct. (b) What is the size of the image? Correct: Your answer is correct. 18 mm (c) Is the image upright or inverted? inverted upright Correct: Image is upright. Your answer is correct. (d) Is the image real or virtual? real virtual Correct: Image is virtual. Your answer is correct. (e) What is the angular magnification if the lens is close to the eye? (Assume the user's near point to be 25 cm.) I ONLY NEED HELP WITH THE LAST ONE!!!!

Explanation / Answer

solution:

       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)

by putting the value
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 = 2.08tion:

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