An arrow 2.10 cm long is located 76.0 cm from a lens that has a focal length f=

Question

An arrow 2.10 cm long is located 76.0 cm from a lens that has a focal length f= 29.5 cm Suppose, instead, that the arrow lies along the principal axis, extending from 75.0 cm to 77.1 cm from the lens, as indicated in the figure (b). What is the longitudinal magnification of the arrow, defined as ? (Hint: Use the thin-lens equation to locate the image of each end of the arrow.) same figure as here http://www.chegg.com/homework-help/questions-and-answers/an-arrow-200-cm-long-is-located-842cm-from-a-lens-which-has-a-focal-length-f-285-cma-if-th-q695139

Explanation / Answer

For left end:- Object distance u1 = -77.1 cm Let v1 = image distance v1 = u1f/(f + u1) = -77.1 * 29.5/(29.5 - 77.1) = 47.78 cm For right end:- Object distance u2 = -75 cm v2 = u2f/(f + u1) = -75 * 29.5/(29.5 - 75) = 48.62 cm Length of image = v2 - v = 48.62-47.78 = 0.84 cm Length of object = u2 - u1 = -77.1 - (-75) = 2.1 cm Li/L0 = .84/2.1 = 0.4

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.