Copy of Lecture-5-Optics

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