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### Multiple Lens Systems

When working with a multiple lens system, it is essen tial to first calculate the position of the image formed by the first lens. Only after locating the first image is it possible to calculate the vergence of light as it reaches the second lens. This is the method by which any number of lenses can be analyzed. Always remember to locate the image formed by the first lens and use it as the object for the second lens to calculate the vergence of light as it reaches the second lens. Repeat the process for each subsequent lens.

**Question: **Where will the image be formed for an object placed 50cm in front of a +4.00D lens that is separated from a –2.00D lens by 25cm. (Figure 8)

Click on image to enlarge.

**Answer:** First, determine the vergence of the image of the object after it passes through the first lens (+4.00D).

Use the equation U1 + D1 = V1 for the first image, where

u1 = object distance = -50cmU1 = object vergence = 100/u1 = 100/-50 = -2.00D

D1 = Lens 1 power = +4.00DV1 = image vergence = U1 + D1 = -2.00D + (+4.00D) = +2.00D

Therefore, Image 1 focuses at v1 = 100/V1 = 100/+2.00D = +50cm behind Lens 1.

Now, Image 1 becomes Object 2 (I1 = O2).

At Lens 2 (-2.00D), Object 2 is located +50cm behind Lens 1 and +25cm behind Lens 2 because Lens 2 is 25cm from Lens 1.Object 2 has a vergence of U2 = 100/u 2 =100/+25 = +4.00D

Using the formula U 2 + D 2 = V 2 whereU2 = Object 2 vergence = +4.00D

D2 = Lens 2 power = -2.00D

V2 = Image 2 vergence = U2 + D2 = +4.00 + (-2.00) = +2.00D. Therefore, Image 2 focuses at v2 = 100/V2 = 100/+2.00D = +50cm behind Lens 2.

**Question: **Consider an object 10cm in front of a +5.00D lens in air. Light strikes the lens with a vergence of 100/u = 100/-10 = –10.00D. (Figure 9)

Click on image to enlarge.

The image has a vergence of V = U + D = -10.00D + (+5.00D) = -5.00D. In this case, light emerges with a negative vergence, which means the light is still diverging after crossing the lens. No real image is produced. In this case, we have a real object and a virtual image. Now suppose that a +6.00D thin lens is placed 5 cm behind the first lens.

- Will an object be formed?
- If so, what are its characteristics?

Image 1 becomes Object 2 and has a vergence of –5.00D. As the light crosses the 5cm to the second lens, its vergence changes. In order to determine the vergence at the second lens, it is necessary to find the location of the image formed by the first lens. If the first lens does not form a real image, it has a virtual image. As light leaves the first lens, it has a vergence of -5.00D. The same vergence would be produced by an object 20cm away if the first lens were not present. So, as light leaves the second lens, it appears to be coming from an object 20cm to the left of the first lens and 25cm away from the second lens. Therefore, the vergence at the second lens is U2 = 100/u2 = 100/-25cm = –4.00D. When light leaves the second lens, it has a vergence of V2 = U2 + D2 = -4.00D + (+6.00D) = +2.00D forming a real image 50cm (v = 100/V = 100/+2.00D) to the right of the second lens.

**Question: **A +2.00D and a –3.00D lens are separated by 30cm. The final image is 20cm behind the second lens (-3.00D). (Figure 10) Where is the object located?

Click on image to enlarge.

**Answer:** In this case we need to work backwards. Using the formula U2 + D2 = V2 where

v2 = Image 2 distance = +20cm

U2 = Object 2 vergence

D2 = Lens 2 power = -3.00D

V2 = Image 2 vergence = 100/v2 = 100/+20cm = +5.00D

U2 = V2 - D2 = +5.00D – (-3.00D) = +8.00D.

Therefore, the location of Image 1/Object 2 is v1=100/V1 =100/+8.00 = +12.5cm right of lens 2. Next, use the formula U1 + D1 = V1 where

v1 = Image 1 distance from Lens 1 = +30 cm + 12.5cm = +42.5cm

U1 = Object 1 vergence

D1 = Lens 1 power = +2.00D

V1 = Image 1 vergence = 100/v 1 = 100/+42.5 = +2.35

U1 = V1 – D1 = +2.35 – (+2.00D) = +0.35D. Therefore, object 1 is located at u1 = 100/U1 = 100/+0.35D = +285.71cm to the right of the first lens.

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