EP 225 Waves, Optics, and Fields

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1 EP 225 Waves, Optics, and Fields Website: contains Course outline Laboratory instruction Notes Past exams Animation Instructor: Akira Hirose Office Physics 66 phone:

2 Marks Assignments (9 or0) 5% Laboratory 5% Midterm 20% Final 50%

3 Laboratory Brian Zulkoskey, Lab Instructor, Room 5 Physics, ph brian.zulkoskey@usask.ca LABS START OF CLASSES Students should be advised that labs for EP 225 will begin Monday, January 7. For these introductory sessions only, students will meet in the rooms listed below at the indicated times: MON 7 JAN :30 p.m. L2A & B RM 26 PHYSICS WED 9 JAN :30 p.m. L4A & B RM 29 PHYSICS Students should bring a copy of "A Laboratory Manual for Engineering Physics 225.3, REVISED 997". These are available at the Book Store. If a student misses the lab period for which he/she is scheduled, he/she should see Brian Zulkoskey in Room 5 Physics as soon as possible.

4 Lab Schedule

5 Subjects to be covered Geometrical optics: reflection, refraction, mirrors, lenses, aberration, optical instruments Oscillations: mechanical (mass-spring, pendulums, energy tossing) Oscillations: E&M (LC circuits, energy tossing, damped oscillations, forced oscillations) Mechanical waves: waves in string, sound waves, water waves, wave reflection, standing waves E&M waves: LC transmission line, characteristic impedance, wave reflection due to impedance mismatch, radiation of EM waves Wave optics: interference, diffraction, resolving power of optical devices

6 Common questions Water waves increase amplitude as they approach a beach. Why? What determine the velocity of light? Sound wave? How does the police radar measure the speed of a car? The frequency of electromagnetic waves in telecommunication has been steadily increasing. Why is high frequency wave more beneficial? How much power is the earth receiving from the sun and in what form? Lenses of high quality optical instruments are coated with dielectric films. Why? What is the fraction of light power reflected at a glass surface? Soap films, CD, DVD appear colored. Why? Is it due to the same mechanism as prism and rainbow? Why are telescopes for astronomical observation so large? Electron microscopes can see better than optical microscopes. Why? How can CAT (computer assisted tomography) and NMRI (nuclear magnetic resonance imaging) image internal organs? How does the antenna work? What is the basic mechanism of radiation of electromagnetic waves? How does a laser work?

7 Light velocity in vacuum In medium with permittivity ε Light Wave c = = εμ c' m/s Harmonic wave c = fλ, f (oscillations/sec = Hz) is the frequency and λ (m) is the wavelength. Index of refraction c ε n = = > c' ε 0 When light wave enters from air to water (n =.33), the wavelength is shortened by the factor n. The frequency remains unchanged because it is determined by wave sources, e.g., microwave generator, laser, etc. 0 0 = < c εμ 0

8 Efforts to find c Roemer (Danish astronomer) observed change in the rotation period of the moon Io revolving around Jupiter. Based on Roemer s 8 data, Huygens deduced c m/s Bradley (British astronomer) found a fairly accurate estimate of 8 c m/s by aberration effect, shift in the angular location of distant star by 4 Δ θ v/ c where v is the earth velocity 3 0 m/s. Fizeau made the first laboratory measurement of the speed of light using rotating toothed wheel.

9 Light reflection and refraction Light path obeys the law of minimum transit time. Reflection angle = incident angle Snell s law for the refraction angle In the figure, λ λ2 c c2 sinθ = = sinθ sinθ sinθ sinθ sinθ 2 2 c n = = c n 2 2 2

10 Prism and Rainbow Prism and rainbow: splitting of white light into color spectrum. n (blue) > n (red) Colors on CD and DVD are due to something else: diffraction.

11 When light emerges from glass to air, Snell s law becomes Total reflection sinθ sin n 2 g g θ = n = = air θ 2.5 Maximum of is 90 degrees. For incident angles > arcsin(/.5) = 4.8 deg, light is totally reflected. Optical fiber waveguide n θ 2 θ c n = sin n2

12 Example: Find the range of incident angle for total reflection at the vertical surface. Total reflection θ > sin = 50.3, θ = = ( ) θ < = sin.3sin

13 Assume small aperture h << R Reflection is symmetric about the radial line Place an object at axial position o (not zero) and find the image location i. α + θ = γ, θ + γ = β α + β = 2 γ, h h 2h + = o i R 2 + = =, f = focal length o i R f Example: Concave mirror with f = 0 cm. Object at 5 cm, real image at 30 cm. Magnification is m=-i/o=-2 (inverted image). Example: Virtual object distance (negative o) Spherical Mirrors

14 Mirrors (cont) Example: Convex mirror (R, f < 0) f = -5 + = 20 i 5 i = 8.6 cm = 8.6 cm behind the mirror 8.6 m = = O R = -30

Spherical mirrors are subject to spherical aberration for large apertures (top). For a large object distance (e.g., stars), parabolic mirror is ideal for focusing (bottom).

15 Spherical mirrors are subject to spherical aberration for large apertures (top). For a large object distance (e.g., stars), parabolic mirror is ideal for focusing (bottom). Schmidt camera (telescope) corrects spherical aberration by placing a lens corrector. Spherical and Parabolic Mirror

16 Mirror Sign Convention, Magnification i, o > 0 real object and real image distance in front of the mirror o, i < 0 virtual object and image distance behind the mirror R > 0, f = R/2 > 0 concave mirror R < 0, f = R/2 < 0 convex mirror Lateral (or angular) magnification m = -i/o, m > 0 erect image, m < 0 inverted

17 Refraction at a Spherical Boundary Snell s law sinθi = nsin θr For small θ, θi = nθr θi = α + γ, θr + β = γ Then α + nβ = n γ ( ) n + = ( n ) o i R sinθ = nsinθ Example: Find the image of a fish at the center of spherical bowl. n (water) = 4/3. From n + = ( n ) o i R n=4/3 4R 4/3 + = R R i 3 R i = R (at the bowl center) i R Magnification is = = +4/3. Ray tracing confirms the result. o/ n 3 R 4 i r

18 Thin Lens First refraction n + = ( n ) o i' R Second refraction n + = ( n) = ( n ) ( i' < 0 in the case shown) i' + t i R R If t is negligible If in water, n + = ( n), i' < 0 in the case shown i i' + t R + = ( n ) =. Lens maker's formula o i R2 R2 f ng nw = f nw R R

19 Sign Convention for Lenses For lenses, f > 0 converging lens, f < 0 diverging lens R > 0 convex surface, R < 0 concave o > 0 real object distance in front of the lens i > 0 real image distance behind the lens Magnification i/ ni m = o/ n In the case of the fish in a bowl, o= R, i = R, n = 4 / 3, n = and m =+ 4 / 3. o o i

20 Thin Lens = ( n ) f R R 2 R>0 R2<0 f > 0 R>0 R2>0 R< R2 f > 0 R>0 R2>0 R>R2 f < 0 R<0 R2>0 f < 0

21 Examples Example: Design lenses with f = +25 cm and 25 cm given n (glass) =.5. = ( n ) = 0.5 = f R R2 R R2 25 R = 2.5 cm, R2 = (plano convex) R = R2 = 25 cm (symmetric) R = 5 cm, R = 75 cm (meniscus) 2 = 0.5 = f R R2 25 R = 2.5 cm, R2 = R = 25 cm, R = 8.33 cm, etc. 2 Example: When a converging lens of f = +30 cm is immersed in a liquid, it becomes a diverging lens of f = - 30 cm. If n (lens glass) =.5, what is n of the liquid? = ( ng ) =, = fair R R2 30 R R2 5 In liquid, ng nliq = = fliq nliq R R2 30 Then n =.696. liq

22 Physical Meaning of Focusing φ Focusing requires the transit time along OHI be independent of the height h or angle φ. ( o δ ) 2 2 Proof: OH HI + + h n 2 2 τ = + n = + ( i δ) + h c c c c 2 h where δ =. 2 R h h 2 2 h h ( o+ δ) + h = o+ +, ( i δ) + h = i + 2R 2 o 2R 2 i 2 o i h n o i τ = + n + + ( n ) = + n (independent of h) c c 2 o i R c c n where + ( n ) = 0 is used. o i R

23 Example: Two lenses Two lenses, lens-mirror 30 cm -0 cm First lens: + =, i ' = 75 cm 50 i ' 30 Second lens: + =, i = 4 cm 35 i 0 O 50 cm 40 cm 75 4 Magnification: m= mm 2 = = 0.6 (erect, virtual) We will revisit this problem in the section of matrix method. The system matrix is 0 0 i i i = From B = i = 0 (the condition for focusing), i = 4 cm.

24 Lens-Mirror System Lens + =, i = i 20 Mirror + =, i ' = 6.67 (-20 means a virtual object) 20 i ' 0 Lens again + =, i '' = 50.0 (final image) i '' Magnification m = ( 2)( ) = +.0 (erect, real) 33.33

25 Compound lens Two lenses touching: Effective focal length + =, = + f i f f f f 2 eff 2 Two lenses separated by d f H2 f 2 + =, = f d f ' f f ' f f d 2 2 f is NOT the focal length of the compound lens. The focal length is (to be shown in matrix section) d f f f' d fd = +, f f ' = principal plane f f f f f f 2 2

26 Myopia Myopia correction d f = -d Myopia is caused by too short a focal length of the eye lens. Placing a diverging lens can correct it. If the far point of the eye is d,the focal length of the correcting lens is f = d. Let the eye lens focal length be fe and the eye lens retina distance be oe. Without lens, + =. With lens + = +. Then f = d. d o f o f f e e e e

27 Hyperopia Hyperopia correction -d Hyperopia is caused by too long a focal length of the eye lens. Placing a converging lens can correct it. If the near point of the eye is - d, the focal length of the correcting lens should be f = d.

28 Chromatic aberration of lens can be corrected by combining converging and diverging lenses made of different glasses as shown. The focal length is Achromatic Doublet Red: = ( nr ) + ( nr' ) fr R R2 R3 R4 Blue: = ( nb ) + ( nb' ) fb R R2 R3 R4 From f = f, R B = ( n n ) ( n ' n ') B R R B R R2 R3 R4

29 Example: Converging lens made of Crown glass ( n =.505, n =.50) and diverging lens made of Flint glass ( n =.65, n =.630). B R B If focal length 50 mm is needed, determine the radii R, R. Achromatic condition = 0.05, R = 2R R R R R 2 Focal length of blue light (= red light) = , Then R = 22.5 mm, R2 = 50 mm 50 R R2 R2

30 2 2 Achromatic Compound Lens The effective focal length of compound lens is d = + f f f f f 2 = ( n ) + -( n ) d R R2 R3 R4 R R2 R3 R4 f + f2 If d =, the focal length f becomes independent of the wavelength. 2 (Assignment Problem)

31 Matrix Method Refraction h= h' n h n n( φ + γ) = n' ( φ' + γ) φ' = + φ n' R n' In matrix form n φ h φ' n' R γ 0 h' h h = n h n φ' = R φ φ n' R n' Transmission over distance t h φ t h'=h+tφ h' t h h = = T φ ' 0 φ φ

32 Matrix of Thin Lens + = φ φ' = h o i f f φ' = h+ φ and h' = h f φ o h f γ i φ' 0 h' h φ' = φ f Two lenses separated by d 0 0 d 0 f f 2 d d f d d d f f f f f f f2 ff2 f 2 =, = C = + eff 2 + f d 2 f2, f ' f' A = C

33 Single Thin Lens If an object is at o from the first lens and image at i from the second lens, total matrix is 0 i o 0 0 f i oi i o i + 0 f f o = = o o f f f i B = 0 determines the image distance o f i =, lens formula i f o i i Magnification = A = = f o

34 Revisit 30 cm -0 cm O 50 cm 40 cm The system matrix is 0 0 i i i = From B = 0 i = 4 cm A = 0.6 (magnification) Focal length = / C = 5cm

35 Principal Planes (thin compound lens) f B: back focal position f H2 f2 f F: front focal position f: effective focal length f'b If the object distance is relative to H and image distance relative to H2, the formal lens formula still holds + = o i f f'f f fd/f2 H d fd/f f

36 For the compound lens in the previous example, fd 5 40 H2 at = = 20 cm to the left of lens 2 f 30 H at fd f = = 60 cm to the right of lens = 60 cm 0 to the left of lens Then the object distance realtive to H is 0 cm. From + =, we find i = 6 cm to the right of H2 or4 cm to the left of lens 2. 0 i 5

37 How to Locate H2 Plane fd f a f f d a = = = f fd a d f Therefore, the H plane can be found from the intersection as shown.

38 Total lens matrix is Thick Lens R H2 R t ( ng ) n g 0 R 2 n g R n n g g t t R n g ng = 2 ( ng ) t t ( ng ) + ( ng ) R R2 ngrr 2 R2ng The effective focal length is ( n ) 2 g t t d = ( ng ) + = +, cf. = + (compound lens) f R R2 ng RR2 f f2 ng ff2 f f f2 ff2 ft ft H2 at to the left of second vertex. H at to the right of first vertex. fn fn 2 g t f f'

39 Example: A lens 3 cm thick has R=0 cm, R2=-5 cm, n (glass) =.5. Determine its focal length and location of the principal planes for the two surfaces. The lens matrix is 0 0 t ( n ) n 0 R 2 n R n cm = = - (.5 ) cm The focal length is f = = = 7.43 cm. C 0.40 A Paraxial ray from left is focused at f ' = = cm. C The principal plane H2 is at =0.74 cm = 7.4 mm from the vertex to the left Similarly, H is at.43 cm to the right of the first vertex.

40 For light coming from right, the order of matrices is reversed (.5 ) cm = cm The focal length is f = = = 7.43 cm (unchanged) C 0.40 A Paraxial ray from right is focused at f ' = = 5.74 cm. C The principal plane H is at =0.74 cm =.429 cm from the vertex to the right

41 Thick Lens Paraxial Ray Diagram H H cm If object at 0 cm from the first vertex, the image is at 8.33 cm from the second vertex

42 Example: An object is placed at 20 cm in front of the thick lens of the previous example. Determine the final image location. In the lens formula, + =, o is to be measured from H and i from H2. o i f eff o = = 2.43 cm From + = i = 0.7 cm i 7.43 Magnification is m = 0.7 / 2.43 = 0.5 In matrix method, i ' i' 20-2 i' = From B = 0, i' = 0 measured from the second vertex. Magnification is A = = 0.5.

43 Magnifying Glass If an object is placed slightly inside the focal point, an erect virtual image is formed. + = o i f of i = < 0 if o< f o f The average image distance is -25 cm. 25 Magnification is m = > 0. f F F 25

44 Telescope In a telescope, the objective lens forms image of distant object at its focal position. The function of the eyepiece is to magnify it as in magnifying glass except the image is at infinity. The image by the objective lens should be formed near the focal point of the eyepiece. The magnification is m β f o = = α = fe f f o e fo fe α β image at infinity

45 Telescope with Erector (for Rifle) To get erect image as needed in rifle telescope, place another lens in between. The total telescope length of ~30 cm and magnification of 8 or 0 is reasonable. An example is shown. In this example, the erector has magnification of -2 and the total magnification is +8. (homework)

46 Microscope In a microscope, an object is placed slightly beyond the focal length of the objective lens to form a magnified real intermediate image at the end of the microscope tube of length L. The function of the eyepiece is to magnify it as in magnifying glass. Magnification is m = L f o 25 all in cm f e

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : , 1 O P T I C S 1. Define resolving power of a telescope & microscope and give the expression for its resolving power. 2. Explain briefly the formation of mirage in deserts. 3. The radii of curvature of