A converging lens has a focal length of 25 cm. Locate the images for the followi

Question

A converging lens has a focal length of 25 cm. Locate the images for the following object distances if they exist. (Enter 0 in the q and M fields if no image exists.)
(a) 25.00 cm
q = Correct: Your answer is correct.
m = Correct: Your answer is correct.
Select all that apply to part (a).
virtual inverted real upright no image Incorrect: Your answer is incorrect.



(b) 6.25 cm
q = Incorrect: Your answer is incorrect.
m = Incorrect: Your answer is incorrect.
Select all that apply to part (b).
upright virtual no image real inverted Correct: Your answer is correct.



(c) 50.00 cm
q = Incorrect: Your answer is incorrect.
m = Incorrect: Your answer is incorrect.
Select all that apply to part (c).
real upright virtual no image inverted Correct: Your answer is correct.

Explanation / Answer

(a) The object is at the focal point of the lens and therefore light from the object emerges from the lens as a parallel beam. There is therefore no image. (b) The lens equation is 1/u + 1/v = 1/f where u is the object distance, v is the image distance and f is the focal length. The convention I use is that distances to real objects and images are positive while distances to virtual objects and images are negative. The focal length of a convex lens is positive and the focal length of a concave lens is negative. 1/2.5 + 1/v = 1/11 1/v = 1/11 - 1/2.5 = -0.309 v = -3.24 cm The image is virtual and upright. Magnification = v/u = 3.24/2.5 = 1.3 (c) 1/u + 1/v = 1/f 1/44 + 1/v = 1/11 1/v = 1/11 - 1/44 = 0.068 v = 14.7 cm The image is real and inverted. magnification = v/u = 14.7/44 = 0.33

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