Consider Refraction at Spherical Surfaces: Starting point for the development of lens equations Vast majority of quality lenses that are used today have segments containing spherical shapes. The aim is to use refraction at surfaces to simultaneously image a large number of object points which may emit at different wavelengths. (Point V (Vertex
sO = SV (object distance( si = VP (image distance( θi - Angle of incidence θt - Angle of refraction θr - Angle of reflection
The ray SA emitted from point S will strike the surface at A, refract towards the .normal, resulting in the ray AP in the second medium (n2( and strike the point P
Using spherical (convex( surfaces for imaging and focusing Object focus
i( Spherical waves from the object focus refracted into plane waves. Suppose that a point at fo is imaged at a point very far away (i.e., si = ∞(. so ≡ fo = object focal length
then ⇒
n1 n2 n2 − n1 + = so ∞ R n1 n2 − n1 = fo R
⇒
n1 fo = R n2 − n1
ii( Plane waves refracted into spherical waves. Suppose now that plane waves (parallel rays( are incident from a point emitting light from a (.point very far away (i.e., so = ∞
when so = ∞ ⇒
n1 n2 n2 − n1 + = ∞ si R
n2 n2 − n1 = si R
⇒
n2 f i ≡ si = R n2 − n1
Diverging rays revealing a virtual image point using concave spherical surfaces. Virtual image point
n2 f i ≡ si = − R n2 − n1
Signs of variables are .important
R<0 fi < 0 si < 0
Parallel rays impinging on a concave surface. The refracted rays diverge and appear to emanate from the virtual focal point Fi. The image is therefore virtual since rays are diverging .from it
A virtual object point resulting from converging rays. Rays converging from the left strike the concave surface and are refracted such that they are parallel to the optical axis. An object is virtual when the rays converge toward it. so < 0 here.
when si = ∞ ⇒
n1 n2 n2 − n1 + = so [ si ] ∞ −R
n1 n2 − n1 = so −R
⇒
n1 f o ≡ so = − R <0 n2 − n1
S (a(
(b (
(c(
As the object distance so is gradually reduced, the conjugate image point P gradually changes from .real to virtual The point Pâ&#x20AC;&#x2122; indicates the position of the virtual image point that would be observed if we were standing in the glass .medium looking towards S
Lateral Magnification
.We will use virtual image points to locate conjugate image points
The Lens Makerâ&#x20AC;&#x2122;s Formula
Newtonian form of the lens equation S2
yo S1
⇒
xo + f − f 1 = f ( xo + f ) xi + f
f2 ⇒ f+ = xi + f xo
⇒
f
( xo + f ) = x xo
i
+f
⇒ xo xi = f 2
:Newtonian Form . xo > 0 if the object is to the left of Fo . xi > 0 if the image is to the right of Fi
The result is that the object ⇒ and image must be on the opposite sides of their .respective focal points
:Define Transverse (or Lateral( Magnification
yi si xi + f f 2 / xo + f f ( f / xo + 1) xo MT ≡ =− =− =− =− ( xo + f ) xo yo so xo + f xo + f =−
x f =− i xo f
Tracing a few key rays through a positive and negative lens
Transverse and Longitudinal Magnification
. m > 0 ⇒ Erect image and m < 0 ⇒ Inverted image All real images for a thin lens .will be inverted
dx ML ≡ i dxo
and
f2 xi = xo
f2 ⇒ M L = − 2 = − M T2 < 0 xo
This implies that a positive dxo corresponds to a negative dxi and vice versa. In other words, a finger pointing toward the lens is imaged pointing away .from it as shown on the next slide
The number-2 ray entering the lens parallel to the central axis limits the image .height
:Image orientation for a thin lens The transverse magnification (MT( is different (.from the longitudinal magnification (ML
Image forming behavior of a thin positive lens. f 2f
f
2f
Location of focal lengths for converging and diverging lenses
nl nlm = >1 nm
nl nlm = <1 nm
a( The effect of placing a second lens L2 within the focal length of a( positive lens L1. (b( when L2 is positive, its presence adds convergence to the bundle of rays. (c( When L2 is negative, it adds divergence to the .bundle of rays
Two thin lenses separated by a distance smaller than either focal length.
Note that d < si1, so that the object for .Lens 2 (L2( is virtual
Note the additional convergence caused by L2 so that the final image is closer to the object. The addition of ray 4 enables the final image to be .located graphically
.Note that d > si1, so that the object for Lens 2 (L2( is real
Fig. 5.30 Two thin lenses separated by a distance greater than the sum of their focal lengths. Because the intermediate image is real, you could start with point Piâ&#x20AC;&#x2122; and treat it as if it were a real object point for L2. Therefore, a ray from Piâ&#x20AC;&#x2122; through Fo2 would arrive at P1.
1 1 1 so1 f1 = − , si1 = si1 f1 so1 so1 − f1 so 2 = d − si1
so 2 < 0 (virtual ( so 2 > 0 (real (
For the compound lens system, so1 is the object distance and si2 is the .image distance
( 1 1 1 so 2 f 2 d − si1 ) f 2 = − , si 2 = = si 2 f 2 so 2 so 2 − f 2 ( d − si1 − f 2 ) The total transverse magnification (MT( is given by
M T = M T 1M T 2
f 2 so1 f1 f2d − so1 − f1 = so1 f1 d − f2 − so1 − f1
si1 si 2 f1si 2 = = − − so1 so 2 d ( so1 − f1 ) − so1 f1
For this two lens system, let’s determine the front focal length (ffl( f1 and .the back focal length (bfl( f2 .Let si2 → ∞ then this gives so2 → f2 so2 = d – si1 = f2 ⇒ si1 = d – f2 but
1 1 1 1 1 = − = − S o1 si 2→∞ f1 si1 f1 d − f 2
⇒
ffl = [ so1 ] si 2→∞
f1 ( d − f 2 ) = d − ( f1 + f 2 )
,From the previous slide, we calculated si2. Therefore, if so1 → ∞ we get
f 2 d − f 2 f1 f 2 ( d − f1 ) bfl = si 2 = = d − f 2 − f1 d − ( f 2 + f1 ) for
f 2 f1 d → 0, bfl = ffl = = f ef f 2 + f1
⇒
1 1 1 = + f ef f1 f 2
”fef = “effective focal length
Suppose that we have in general a system of N lenses whose thicknesses are .small and each lens is placed in contact with its neighbor :Then, in the thin lens approximation
N â&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Ś3 2
1
1 1 1 1 1 = + + ... f ef f1 f 2 f 3 f N
Fig. 5.31 A positive and negative thin lens combination for a system having a large spacing between the lenses. Parallel rays impinging on the first lens enable the position of the bfl.
Example B
Example A
Example A: Two identical converging (convex( lenses have f1 = f2 = +15 cm and separated by d = 6 cm. so1 = 10 cm. Find the position and magnification .of the final image
1 1 1 + = so1 si1 f1
⇒ si1 = -30 cm at (O’( which is virtual and erect Then so2 = |si1| + d = 30 cm + 6 cm = 36 cm
1 1 1 + = so 2 si 2 f 2
⇒ ’si2 = i’ = +26 cm at I
Thus, the image is real and .inverted
The magnification is given by
M T = M T 1M T 2
si1 si 2 − 30 26 = − = − − − = −2.17 so1 so 2 10 36
Thus, an object of height yo1 = 1 cm has an image height of yi2 = -2.17cm Example B: f1= +12 cm, f2 = -32 cm, d = 22 cm An object is placed 18 cm to the left of the first lens (so1 = 18 cm(. .Find the location and magnification of the final image
1 1 1 + = so1 si1 f1
1 1 1 + = so 2 si 2 f 2 M T = M T 1M T 2
⇒ si1 = +36 cm in back of the second lens, and thus creates a virtual object for the .second lens
Image is real and Inverted
so2 = -|36 cm – 22 cm| = -14 cm
⇒ si2 = i’ = +25 cm; The magnification is given by
s s 36 25 = − i1 − i 2 = − − = −3.57 so1 so 2 18 − 14
Thus, if yo1 = 1 cm this gives yi2 = -3.57 cm