Two-lens systems. The stick figure O (the object) stands on the common central a

Question

Two-lens systems. The stick figure O (the object) stands on the common central axis of two thin, symmetric lenses. Lens 1 is mounted within the boxed region closer to O, which is at object distance do 1. Lens 2 is mounted within the farther boxed region, at distance d.

The table refers to the focal distances f1 and f2 for lens 1 and lens 2 respectively, the image distance di 2 for the final image produced by lens 2 (the final image produced by the system) and the overall lateral magnification mtot for the system. All distances are in centimeters. Fill in the missing information, including signs.

* signs not given - you figure out what's appropriate!

Explanation / Answer

According to given set of values,

di2 = image distance
mtot = magnification
Real is Positive sign convention.

Part-1:Image formation of a Diverging Lens:

1/d0 + 1/di1 = 1/f1

1 / 7.5 + 1 / di1 = 1 / 5.5

di1 = 20.625cm

mt = 20.625 /7.5 = 2.75

Virtual image 20.625 cm. left of the diverging lens with magnification 2.75 .

Part-2: Image formation of a converging Lens:

Virtual image becomes a real object, left of the coverging lens

do2 = di1 + d

do2 = 20.625 +12 = 32.625

Now, 1 / 32.625 + 1 / di2 = - 1 / 6.4

di2 = - 0.187cm.

mto = -0.187/32.625 = -0.0057

Real inverted image -0.187cm. left of the Converging lens.

Overall Magnification = 2.75 * (-0.187) / 32.625

mtot = 0.01575.

I think you understand my explanation, and hope you got your solution and glad I can help .Rate me! if you like the solution.

Thank you .

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