Problem 24.28 A 15-cm-focal-length converging lens is 19 cm to the right of a 6.

Question

Problem 24.28 A 15-cm-focal-length converging lens is 19 cm to the right of a 6.0-cm-focal-length converging lens. A 2.3-cm-tall object is distance L to the left of the first lens.

Part A For what value of L is the final image of this two-lens system halfway between the two lenses? Express your answer to two significant figures and include the appropriate units.

Part B What is the height of the final image? Express your answer to two significant figures and include the appropriate units.

Part C What is the orientation of the final image? What is the orientation of the final image?

upright / inverted

Explanation / Answer

focal length of right lens fr = 15 cm

focal length of left lens fl= 6 cm
height of the object h = 2.3 cm
a)
from given problem , image distance i2 =-19/2 = -9.5 cm (for converging lens)
the thin lens equation is
1/fr =1/o2 + 1/i2
O2 = fri2(i2-fr) plug in the values
O2 = 5.82
Now the image is formed at the 5.82 cm left from second lens.
therefore, the distance between object and image from the first lens.
i1= 19 cm - 5.82 cm = 13.18cm
from the thin lens equation ,
1/fl =1/o1 + 1/i1
O1 = fl i1/(i1-fl)   
Plug in the values to get
O1 = 11 cm = L
B) and D)
We knows that total magnification M = m1xm2
=(-i1/o1)(-i2/o2) plug in values to get
M = -1.95
final image height h' = hM = (2.3)(-1.7)
= -4.49 cm   

The image formed is inverted.

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