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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.

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

P = P_{1}+ P_{2}

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?

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

F_{1} = *d*_{1}*/**4*

=*48 cm**/**4*

= 12 cm

=

= 12 cm

and for the diverging lens

F_{2} = *d*_{2}*/**4*

=*32 cm**/**4*

= 8 cm

=

= 8 cm

We take the focal length of the diverging lens as negative. Thus, we write F_{2} = - 8 cm.

Since these two lenses are placed in contact, we obtain for the total optical power of this system:

P = P_{1} + P_{2}

=*1**/**F*_{1} + *1**/**F*_{2}

=*1**/**0.12 m* + *-1**/**0.08 m*

=*2**/**0.24 m* - *3**/**0.24 m*

= -*1**/**0.24 m*

≈ - 4.17 dipotres

=

=

=

= -

≈ - 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

P = P_{1} + P_{2} - d × (P_{1} × P_{2})

where d is the distance (in metres) between the two lenses and P_{1} and P_{2} are the individual optical powers.

Two lenses are placed at the positions shown in the figure.

What is the optical power of this system of lenses?

First, let's work out the respective focal lengths. From the figure, you can see that the focal length of the converging lens is F_{1} = 20 cm / 2 = 10 cm = 0.1 m and for the diverging lens, the focal length is F_{2} = 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

P = P_{1} + P_{2} - d × (P_{1} × P_{2} )

=*1**/**F*_{1} + *1**/**F*_{2} - d × (*1**/**F*_{1} × *1**/**F*_{2} )

=*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

=

=

=

=

=

= -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.

Enjoy the "Power of a Combination of Lenses" physics lesson? People who liked the "Power of Lenses. The Human Eye lesson found the following resources useful:

- Combination Feedback. Helps other - Leave a rating for this combination (see below)
- Optics Physics tutorial: Power of Lenses. The Human Eye. Read the Power of Lenses. The Human Eye physics tutorial and build your physics knowledge of Optics
- Optics Revision Notes: Power of Lenses. The Human Eye. Print the notes so you can revise the key points covered in the physics tutorial for Power of Lenses. The Human Eye
- Optics Practice Questions: Power of Lenses. The Human Eye. Test and improve your knowledge of Power of Lenses. The Human Eye with example questins and answers
- Check your calculations for Optics questions with our excellent Optics calculators which contain full equations and calculations clearly displayed line by line. See the Optics Calculators by iCalculator™ below.
- Continuing learning optics - read our next physics tutorial: Features of Light

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