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Physics Lesson 12.10.2 - Power of a Combination of Lenses
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Welcome to our Physics lesson on Power of a Combination of Lenses, this is the second lesson of our suite of physics lessons covering the topic of Power of Lenses. The Human Eye, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Power of a Combination of Lenses
When two lenses are in contact, the focal length of this system is calculated by the formula
(Here, we do not consider the lenses' thickness, which in reality makes them slightly displaced from each other.)
Therefore, the optical power of such a system of lenses, is
Example 1
What is the power of the system composed by two lenses as those shown in the figure, if they are placed in contact with each other?
Solution 1
The given values represent the distance between the centres of curvatures in each lens. Therefore, we must divide them by 4 to obtain the respective focal lengths. Thus, we have for the converging lens
= 48 cm/4
= 12 cm
and for the diverging lens
= 32 cm/4
= 8 cm
We take the focal length of the diverging lens as negative. Thus, we write F2 = - 8 cm.
Since these two lenses are placed in contact, we obtain for the total optical power of this system:
= 1/F1 + 1/F2
= 1/0.12 m + -1/0.08 m
= 2/0.24 m - 3/0.24 m
= - 1/0.24 m
≈ - 4.17 dipotres
When two lenses are not in contact, we must also consider the distance between them in order to calculate the optical power of this system. The formula of optical power in this case becomes
where d is the distance (in metres) between the two lenses and P1 and P2 are the individual optical powers.
Example 2
Two lenses are placed at the positions shown in the figure.
What is the optical power of this system of lenses?
Solution 2
First, let's work out the respective focal lengths. From the figure, you can see that the focal length of the converging lens is F1 = 20 cm / 2 = 10 cm = 0.1 m and for the diverging lens, the focal length is F2 = 12 cm / 2 = 6 cm = 0.06 m (we must write - 0.06 m as it is the focal length of a converging lens).
Also, we can determine the distance between the two lenses. It is d = 10 cm + 8 cm + 6 cm = 24 cm = 0.24 m. Thus, the optical power of this system is
= 1/F1 + 1/F2 - d × (1/F1 × 1/F2 )
= 1/0.1 m + -1/0.06 m - 0.24 m × (1/0.1 m × -1/0.06 m)
= 3 - 5/0.3 - 0.24 × (-1/0.006)
= -2/0.3 + 0.24/0.006
= 20/3 + 240/6
= -6.67 + 40
= 33.33 dioptres
You have reached the end of Physics lesson 12.10.2 Power of a Combination of Lenses. There are 3 lessons in this physics tutorial covering Power of Lenses. The Human Eye, you can access all the lessons from this tutorial below.
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- Continuing learning optics - read our next physics tutorial: Features of Light
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