I can\'t seem to find any solutions to this problem online and I\'m extremely st

Question

I can't seem to find any solutions to this problem online and I'm extremely stuck. I beleive the concept is that the light from the object passes through the lens, is then reflected by the mirror and then the light passes through the lens again to form the image. But none of my math is working. Please walk through your solution step by step.

Thank You

Problem:

The lens and mirror in the figure below are separated by d = 1.00 m and have focal lengths of +74.4 cm and -54.9 cm, respectively. An object is placed p = 1.00 m to the left of the lens as shown.

(a) Locate the final image, formed by light that has gone through the lens twice.

distance:

location: To the left of the lens.

(b) Determine the overall magnification of the image.:

(c) State whether the image is upright or inverted.:

Explanation / Answer

1/u1 + 1/v1 =1/f1

u1 = 1m

f1 = 74.4cm = 0.744 m

1/v1 = 1/f1 - 1/u1

v1 = 2.90625 m

m1 = -v1/u1 = -3.90625

the object of the mirror = 2.90625 -1 = 1.90625m to the right of the mirror so object is virtual . u2 = -1.90625

1/v2 = 1/f2 - 1/u2

1/v2 = 1/(-0.549) - 1/(-1.90625)

v2 = -0.77 m

m2 = -v2/u2 = -0.4045

thi image formed the mirror is 1+0.77 = 1.77 m to the right of lens

u3 = 1.77 m

1/v3 = 1/f1 - 1/u3

v3 = 1.2835m to the left of lens

m3 = -v3/u3 = -0.725

M = m1*m2*m3

M = -1.1456

part c ) M<0 so inverted

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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