Question

2. The separation between two thin lenses (fl = +20 cm, left, and f2 -15 cm right) is 30 cm. An object is located 40 cm away on the left of the positive lens. (a) Find the location and magnification of the final image and describe the image (b) Find the matrix of this two-lens system, and use the matrix to locate the final image. (c) If the object is located at infinity, where is the final image?

Answer 1

( a )

The formula for the effective focal length ?f of two thin lenses with focal lengths ?1 and ?2 separated by distance ? to be

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However, I can't seem to find how ? is defined. Is it the distance from the first lens to the final focal point or the distance from the second lens to the final focal point? Or neither?

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

It is the distance from the image plane to the rear principal plane. You can find the location of this plane by projecting the image ray backward through the system to where it crosses the projection of the object ray. This is sometimes also referred to as the effective focal length (EFL) of the system and is true for both simple as well as complicated systems. The distance from the rear lens to the image plane is simply the back focal distance (BFD). The difference between the EFL and BFD can be found by the formula:

delta = -(d/n)*(f/f1) = -(EFL-BFD); where n=1 in air

BED FEL - - - -

Thus the position of the first image

Thin lens equation:

$\frac{1}{f_{1}}=\frac{1}{S_{01}}+\frac{1}{S_{i1}}$

Solve for

(三) - S

S = (sem com) = 40cm

This is a positive number, so the image is to the right of the first lens.

Calculate the second object distance,

cm( -10 cm )

So2 = d - Si = 30 - 40 = 10

calculate position of second image

Thin lens equation:

11_1 Soz * Siz fz

o 4) = as

cm

(三 ) - 30

This is a positive number, so the image is to the right of the second lens.

Magnification of the first lens:

Si mti = 40 40 So =

Magnification of the second lens:

Si2 mt2 = 30 -10

Total magnification

mtimt2 = -1*3= -3

( B )

System Matrix: Two Thin Lenses

The position of the image formed by a pair of thin lenses can be found by the matrix method. The general matrix method involves multiplying a vector form of the incident vergence successively by matrices representing (1)the refraction by the first lens, (2) the translation to the second lens, and (3) the refraction by the second lens. If the system matrix is calculated, it can be used to directly multiply the incident vergence to obtain the exit vergence. From that exit vergence, the image distance is calculated.

100 1 p, 1 0 1 Lo 1][á 1][0 P.Tv] [kv] Divide out 1][1] [k ] kto get V. scale factor Calculation of the system matrix gives this form.

The angular magnification of the system The equivalent power of the system

( c ) If the object is at an infinitesimal distance then the final image appears at the resultant focal length of the combined system i.e

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