Two converging lenses ( f 1 = 9.00 cm and f 2 = 6.00 cm) are separated by 18.0 c

Question

Two converging lenses (f1 = 9.00 cm and f2 = 6.00 cm) are separated by 18.0 cm. The lens on the left has the longer focal length. An object stands 11.2 cm to the left of the left-hand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, (d) is the final image upright or inverted and (e) is it larger or smaller?

(a) Number _____ Units _____

(b) Number _____ Units _____

(c) The final image is _____ (real or virtual)

(d) The final image is _____ (upright or inverted)

(e) The final image is _____ (larger or smaller)

Explanation / Answer

u1 = (9cm + 11.2 cm) = 20.2 cm =0.202m
f1 = 0.09 m
separation = 0.18 m
f2 = 0.06 m

A
For the 1st lens:
v1 = 1 / (1/f1 - 1/u1)
   = 1 / (1/0.09 - 1/0.202 )
   = 1/6.16
   = 0.162 m (Here it is "q1", to right of lens 1)
  
For the 2nd lens:
u2 = sep - v1
   = 0.18 - 0.162
   = 0.018 m (the intermediate image is to the left of lens 2)
  
v2 = 1 / (1/f2 - 1/u2)
   = 1 / (1/0.06 - 1/0.018)
   = 1 / (16.66-55.55)
   = 1 / 1.11
   = 0.901 m ("q2", to right of lens 2)
  
Magnification = v1v2/(u1u2)
                 = (0.162*0.901)/(.202*0.018)
                = 0.146/0.0036
                = 40.55

h'/h = -(q/p)

h' = height of the object's reflection =
h = height of the object = 0.112
q = distance of object from mirror = 0.202 m
p = distance of reflection in mirror = 0.901

h'/0.112 = -(0.202/.901)

h' = -0.025 m

(a) the final image relative to the lens on the right is 0.901 m

(b) Overall magnification is 40.55

(c) The final image is real

(d) The final image is inverted

(e) Final image is smaller.

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