SE9 36. P.080 An object is originally at the xi = 0 cm position of a metersti fi

Question

SE9 36. P.080 An object is originally at the xi = 0 cm position of a metersti fixed at the position 31.0 cm. Then we gradually slide the object to the position xf ck located on the x axis. A converging lens of focal length 30.0 crnis 10.3 cm (a) Find the location x of the object's image as a function of the object position x. ( Assume x and x are in cm.) x'=1 19.7 (b) Describe the pattern of the image's motion with reference to a graph or a table of values. (Do this on paper. Your instructor may ask you to turn in this work.) (c) As the object moves 10.3 cm to the right, how far does the image move?

Explanation / Answer

given xi = 0 cm

f = 30 cm, xf = 31 cm

x" = 10.3 cm

a. location of image = x'

now, object distance = -(xf - x) = u

v = x' - xf

hence

1/v - 1/u = 1/f

1/(x' - xf) = 1/f - 1/(xf - x)

x' - xf = f(xf - x)/(xf - x - f)

x' = xf + f(xf - x)/(xf - x - f)

hence

x' = 31 + 30(x - 31)/(x - 1)

b. for 0 < x < 1 cm

x' > 0 and increases with increase in x

for x = 1 cm

x' = inf

for x > 1 cm

x' < 0 and |x'| decrease with increase in x

c. for x = xi

x' = 961 cm

for x = x"

x' = -35.7741 cm

hence the image moves 996.774 cm to the left

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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