An object is placed 16.4 cm from a first converging lens of focal length 11.8 cm

Question

An object is placed 16.4 cm from a first converging lens of focal length 11.8 cm. A second converging lens with focal length 12 cm is placed 10 cm to the right of the first converging lens. (Take the direction to the right to be positive.)
(a) Find the position q1 of the image formed by the first converging lens.
cm

(b) How far from the second lens is the image of the first lens?
cm

(c) What is the value of p2, the object position for the second lens?
cm

(d) Find the position q2 of the image formed by the second lens.


(e) Calculate the magnification of the first lens.
M1 =

(f) Calculate the magnification of the second lens.
M2 =

(g) What is the total magnification for the system?
Mtotal =
(h) Is the final image real or virtual? Is it upright or inverted?

inverted
real
virtual
no image
upright

Explanation / Answer

Given Focal length of first converging lens f 1 = 11.8 cm Focal length of second converging lens f 2 = 12 cm Object distance of first lens   p1 = 16.4 cm Distance between two lenses   d = 10 cm a) Image formed by first converging lens is           1 / p1 + 1 / q 1   = 1/ f 1                         1 / q 1   = 1 / f 1 - 1 / p 1                                     = 1 / 11.8 cm - 1 / 16.4 cm                                     = 0.02377                               q 1 = 42.069 cm   _____________________________________________ b) Disatnce between image of first lens and second lens is                             p2 = q 1 - d                                  = 42.069 cm - 10 cm                                  = 32.069 cm _______________________________________________ c) object position for the second lens                         p2 = - 32.069 cm _______________________________________________ d) Image formed by the second lens.             1 / p2 + 1 / q 2   = 1 / f 2                           1 / q 2   = 1 / f 2 - 1 / p2                                        = 1 / 12 cm  + 1 / 32.069 cm                                q 2   = 8.7324 cm   _______________________________________________ e) Magnification of the first lens                      m 1 = - q 1 / p 1                             = - ( 42.069 cm ) / ( 16.4 cm )                             = - 2.5651 _______________________________________________ f) Magnification of second lens is                    m 2 = - q 2 / p 2                           = - 8.7324 / 32.069                           = - 0.2723 _______________________________________________ g) Total magnification for the system is                  M =  m 1 m2                        = ( - 2.5651 ) ( -0.2723 )                        = 0.6984 Final image is virtual image .

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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