Topic 3: Operation of Simple Lens Aim: Covers imaging of simple lens - PDF document

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Topic 3: Operation of Simple Lens Aim: Covers imaging of simple lens using Fresnel Diffraction, resolu- tion limits and basics of aberrations theory. Contents: 1. Phase and Pupil Functions of a lens 2. Image of Axial Point 3. Example of Round Lens 4. Diffraction limit of lens 5. Defocus 6. The Strehl Limit 7. Other Aberrations O P T I C D S E G I R L O P P U A P D S Properties of a Lens -1- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Ray Model Simple Ray Optics gives f Image Object u v Imaging properties of 1 u + 1 v = 1 f The focal length is given by � 1 � 1 + 1 f = ( n − 1 ) R 1 R 2 For Infinite object Phase Shift Ray Optics gives Delta Fn f Lens introduces a path length difference, or PHASE SHIFT. O P T I C D S E G I R L O P P U A P D S Properties of a Lens -2- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Phase Function of a Lens δ 1 δ 2 h R 2 R 1 n P P 0 1 ∆ With NO lens, Phase Shift between , P 0 → P 1 is where κ = 2 π Φ = κ∆ λ with lens in place, at distance h from optical,   Φ = κ  δ 1 + δ 2 + n ( ∆ − δ 1 − δ 2 )  � �� � � �� � Air Glass which can be arranged to give Φ = κ n ∆ − κ ( n − 1 )( δ 1 + δ 2 ) where δ 1 and δ 2 depend on h , the ray height. O P T I C D S E G I R L O P P U A P D S Properties of a Lens -3- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Parabolic Approximation Lens surfaces are Spherical , but: If R 1 & R 2 ≫ h , take parabolic approximation δ 1 = h 2 δ 2 = h 2 and 2 R 1 2 R 2 So that � 1 � Φ ( h ) = κ∆ n + κ h 2 + 1 2 ( n − 1 ) R 1 R 2 substituting for focal length, we get Φ ( h ) = κ∆ n − κ h 2 2 f So in 2-dimensions, h 2 = x 2 + y 2 , so that Φ ( x , y ; f ) = κ∆ n − κ ( x 2 + y 2 ) 2 f the phase function of the lens. Note: In many cases κ∆ n can be ignored since absolute phase not usually important. The difference between Parabolic and Spherical surfaces will be con- sidered in the next lecture. O P T I C D S E G I R L O P P U A P D S Properties of a Lens -4- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Pupil Function The pupil function is used to define the physical size and shape of the lens. p ( x , y ) ⇒ Shape of lens so the total effect of the lens of focal length is p ( x , y ) exp ( ı Φ ( x , y ; f )) For a circular lens of radius a , if x 2 + y 2 ≤ a 2 p ( x , y ) = 1 = 0 else The circular lens is the most common, but all the following results apply equally well for other shapes. Pupil Function of a simple lens is real and positive, but it will be used later to include aberrations, and will become complex. O P T I C D S E G I R L O P P U A P D S Properties of a Lens -5- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Fresnel Image of Axial Point We will now consider the imaging of an axial point using the Fresnel propagation equations from the last lecture. Consider the system. u (x,y) u (x,y) f 2 0 z z 1 P P’ P P 0 1 1 0 2 In plane P 0 we have that u 0 ( x , y ) = A 0 δ ( x , y ) so in plane P 1 a distance z 0 , we have that, u ( x , y ; z 0 ) = h ( x , y ; z 0 ) ⊙ A 0 δ ( x , y ) = A 0 h ( x , y ; z 0 ) where h ( x , y ; z ) is the free space impulse response function. If we now assume that the lens is thin , then there is no diffraction between planes, P 1 and P ′ 1 , so in P ′ 1 we have u ′ ( x , y ; z 0 ) = A 0 h ( x , y ; z 0 ) p ( x , y ) exp ( ı Φ ( x , y ; f )) so finally in P 2 a further distance z 1 we have that u 2 ( x , y ) = u ′ ( x , y ; z 0 ) ⊙ h ( x , y ; z 1 ) = A 0 h ( x , y ; z 0 ) p ( x , y ) exp ( ı Φ ( x , y ; f )) ⊙ h ( x , y ; z 1 ) O P T I C D S E G I R L O P P U A P D S Properties of a Lens -6- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Take the Fresnel approximation, where � � h ( x , y ; z ) = exp ( ı κ z ) ı κ 2 z ( x 2 + y 2 ) exp ı λ z and Φ ( x , y ; f ) = κ n ∆ − κ 2 f ( x 2 + y 2 ) so we get that � � u ( x , y ; z 0 ) = A 0 exp ( ı κ z 0 ) ı κ ( x 2 + y 2 ) exp ı λ z 0 2 z 0 then by more substitution, A 0 exp ( ı κ ( z 0 + n ∆ )) u ′ ( x , y ; z 0 ) = p ( x , y ) ı λ z 0 � 1 � � � ı κ − 1 ( x 2 + y 2 ) exp 2 z 0 f Finally!, we can substitute and expand to get 1 � �� � A 0 u 2 ( x , y ) = exp ( ı κ ( z 0 + z 1 + n ∆ )) λ 2 z 0 z 1 2 � �� � � � ı κ ( x 2 + y 2 ) exp 2 z 1 3 � 1 � �� � � �� ı κ + 1 − 1 Z Z 2 ( s 2 + t 2 ) p ( s , t ) exp z 0 z 1 f � � − ı κ ( sx + ty ) exp d s d t z 1 � �� � 4 O P T I C D S E G I R L O P P U A P D S Properties of a Lens -7- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Look at the term: 1. Constant, amplitude gives absolute brightness, phase not mea- surable. 2. Quadratic phase term in plane P 2 , no effect on intensity, and usually ignored. 3. Quadratic phase term across the Pupil, depends on both z 0 and z 1 . 4. Scaled Fourier Transform of Pupil Function. If we select the location of plane P 2 such that 1 + 1 − 1 f = 0 z 0 z 1 Note same expression as Ray Optics, then we can write � � ı κ ( x 2 + y 2 ) u 2 ( x , y ) = B 0 exp 2 z 1 � � − ı κ Z Z p ( s , t ) exp ( sx + ty ) d s d t z 1 so in plane P 2 we have the scaled Fourier Transform of the Pupil Function, (plus phase term). Intensity in plane P 2 is thus | u 2 ( x , y ) | 2 g ( x , y ) = � � � � 2 − ı κ Z Z � � B 2 = p ( s , t ) exp ( sx + ty ) d s d t � � 0 z 1 � � Being the scaled power spectrum of the Pupil Function. O P T I C D S E G I R L O P P U A P D S Properties of a Lens -8- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Image of a Distant Object For a distant object z 0 → ∞ and z 1 → f u (x,y) p(x,y) 2 P P’ f z 1 1 P 0 2 Then the amplitude in P 2 becomes, � � ı κ 2 f ( x 2 + y 2 ) u 2 ( x , y ) = B 0 exp � � − ı κ Z Z p ( s , t ) exp f ( sx + ty ) d s d t which being the scaled Fourier Transform of the Pupil function. The intensity is therefore: � � � � 2 − ı κ Z Z � � g ( x , y ) = B 2 p ( s , t ) exp f ( sx + ty ) d s d t � � 0 � � which is known as the Point Spread Function of the lens Key Result Note on Units: x , y , s , t all have units of length m . Scaler Fourier kernal is: � � − ı 2 π λ f ( sx + ty ) exp O P T I C D S E G I R L O P P U A P D S Properties of a Lens -9- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Simple Round Lens Consider the case of round lens of radius a , amplitude in P 2 then becomes, � � − ı κ Z Z u 2 ( x , y ) = ˆ f ( sx + ty ) B 0 exp d s d t s 2 + t 2 ≤ a 2 the external phase term has been absorbed into ˆ B 0 This can be integrated, using standard results (see Physics 3 Optics notes or tutorial solution), to give: � � κ a B 0 a 2 J 1 f x u 2 ( x , 0 ) = 2 π ˆ κ a f x where J 1 is the first order Bessel Function. This is normally normalised so that u 2 ( 0 , 0 ) = 1 , and noting that the output is circularly symmetric, we get that, � � κ a J 1 f r u 2 ( x , y ) = 2 κ a f r where x 2 + y 2 = r 2 . The intensity PSF is then given by, � � 2 � � κ a � � J 1 f x � � g ( x , y ) = 4 � � κ a � f x � � � which is known as the Airy Distribution, first derived in 1835. O P T I C D S E G I R L O P P U A P D S Properties of a Lens -10- Autumn Term C E P I S A Y R H T P M f E o N T

N E U R S E I H T Modern Optics T Y O H F G R E U D B I N Shape of jinc The function jinc ( x ) = 2J 1 ( x ) x 1 jinc(x) 0.8 0.6 0.4 0.2 0 -0.2 -15 -10 -5 0 5 10 15 Similar shape to the sinc () function, except • Zeros at different locations. • Lower secondary maximas Zeros of jinc () located at 3 . 832 1 . 22 π x 0 7 . 016 2 . 23 π x 0 10 . 174 3 . 24 π x 0 13 . 324 4 . 24 π x 0 O P T I C D S E G I R L O P P U A P D S Properties of a Lens -11- Autumn Term C E P I S A Y R H T P M f E o N T