Name: Thin Lenses in Contact
Uploaded: 2019-07-19
Duration: 4 min 6 s
Description: If two or more lenses are in contact, we can treat them as one lens by calculating the effective focal length of the lenses. Watch and learn the formula for...

Question

How is focal length defined for a two lens system, separated by distance d?

Accepted Answer

Let us consider two lenses of two different focal lengths f1 and f2 placed coaxially (in one common axis). If we can find one single lens that can produce the same image at the same place and of the same size as the above setup, then we can state that this one single lens is equivalent to the above combination of lenses. This is what an equivalent lens means.

Let us consider two converging lenses 1 and 2 , of second focal lengths f1 and f2 respectively, and keep them separated by a distance dd as shown in the given figure. A ray AB moving parallel to the common axis of the lens is incident at a height y1 above the axis of the first lens. This ray after refraction through the first lens is directed towards the second principle focal point Q1 of this lens. This deviation a1 produced by the first lens is given by:

a1≈y1f1

where f1=O1Q1 is the second focal length of the first lens.

The emergent ray from the first lens is incident on the second lens at a height y2 above the axis. After refraction through the second lens, the ray finally meets the principle axis at F2, which is the second focal point of the combination. The deviation a2 produced by the second lens is given by

a2≈y2/f2

Where f2 is the second focal length of the second lens.

The conjugate rays AB and CF2 when produced intersect at E2. if now a converging lens be placed at E2H2 having its second focus at F2 and focal length F2=H2F2 , then the parallel ray ABE2 incident at E2 of the lens E2H2 will be refracted along E2CF2 producing the same total deviation aa as that produced by the two lenses together. Hence, the lens of focal length H2F2 placed at E2H2 will be equivalent to the lens combination. The plane E2H2 is called the second principle plane and the point H2, the second principle point. The deviation produced by the equivalent lens is given by

a=y1/f

Now, a=a1+a2

⇒y1/f=y1/f1+y2/f2

⇒y1/f=y1/f1+y1−(y1−y2)/f2 - equation (1)

From the figure, we can write

y1−y2=CG=d.a1=d.y1/f1

Substituting this in the equation (1) we get,

y1/f=y1/f1+y1/f2−dy1/f1f2

⇒1/f=1/f1+1/f2−d/f1f2

Hence the equivalent focal length,

f=−f1f2/{d−(f1+f2)} (m)

Since power (P)=1f(P)=1f

The equivalent power is,

P=1/f1+1/f2−d/f1f2

How is focal length defined for a two lens system, separated by distance d?

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