problem- an object is placed in front of a converging lens at a distance equal t

Question

problem- an object is placed in front of a converging lens at a distance equal to twice the focal length f1 of the lens. On the other side of the lens is a concave mirror of focal length f2 separated from the lens by a distance 2(f1 + f2). Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. Take f1 = 4.5 cm and f2 = 7.9 cm. What are (a) the distance between the lens and the final image and (b) the overall lateral magnification M of the object?

Explanation / Answer

(a). the distance of object measured from a concave mirror is s2 = 2(f1 + f2) - 2f1 = 2f2 first, the light refract from a concave mirror mirror : 1/s2 + 1/s2' = 1/f2 1/(2f2) + 1/s2' = 1/f2 1/s2' = 1/(2f2) s2' = 2f2 the distance of image from a converging lens is s1 = 2(f1 + f2) - s2' = 2(f1 + f2) - 2f2 s1 = 2f1 Let s1' is the distance of final image from a converging lens. 1/s1 + 1/s1' = 1/f1 1/(2f1) + 1/s1' = 1/f1 1/s1' = 1/(2f1) s1' = 2f1 = 2(4.5) s1' = 9.0 cm (b). M = (s1'/s1)(s2'/s2) = (2f1/2f1)(2f2/2f2) M = 1

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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