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Lens Equation – Thin Lens h s’ s f h’ Refraction on Spherical Surface θa=α +φ ; φ = β +θb ⎫ an sinθ a= n bsinθ b n⎬aα + n bβ = ( n b− )φ n a anθ a≅ n b θ b ⎭ h h h tanα = ; tanβ = ; tanφ = s + δ s'−δ R − δ h h h n n n− n ≅α ; β ≅; φ ≅ ⇒a+ b = b a s s' R s s' R Refraction on Spherical Surface n n n− n a+ b = b a Magnification s s' R R positive if C on transmission side; negative otherwise y − y′ tanθ = tanθ = na y nb y′ a b = − s s′ s s′ nasinθ a= n bsin θ b y′ na ′ s esl angallFor sm m = = − y n s tanθ≅ sin θ b Refraction on Spherical Surface n n n− n a+ b = b a . A fish is 7.5Example: Fish bowl s s' R cm from the front of the bowl. y′ n′ s Find the location and magnification of the m = = − a fish as seen by the cat.bowl. Ignore effect of y nb s Radius of bowl = 15 cm. na = 1.33 R negative b O n n n− n n n− n n I s’ a+ b =b a b = b a− a s s' R s' R s s n 1 s′ = b = =6 .− 44cm n− n n 0− . 331 . 33 a b a− a − R s −cm15 7 . 5 cm n′ s 1 .( 33 6− . 44cm ) The fish appears closer and larger. m= − a= − ∗ 1= . 14 nb s 1 7 .cm 5 Lenses • A lens is a piece of transparent material shaped such that parallel light rays are refracted towards a point, a focus: – Convergent Lens Positive f » light moving from air into glass will move toward the normal » light moving from glass back into air will move away from the normal » real focus Negative f – Divergent Lens » light moving from air into glass will move toward the normal » light moving from glass back into air will move away from the normal » virtual focus Lens Equation – thin lens n n n− n n n n− n a b+ b = a b+ a = a b 1s s 1' R 1 2s s 2' R2 For air, na=1 and glass, nb=n, and s2=-s1’. Neglects l. 1 n n − 1 n 1 1 − n + = − + = 1s s 1' R 1 1s' s 2' R 2 Eliminate, s1’, s1 → s, s2’ → s’ 1 1 1⎛ 1⎞ 1 ⇐ Lensmaker qn.e + =()n −1 ⎜ −⎟ = f > 0, converging s s' R⎜ R⎟ f ⎝ 1 2⎠ f < 0, diverging Lensmaker’s Formula The thin lens equation for a lens with two curved sides: 1 ⎛ 1 1 ⎞ =()n −1 ⎜ − ⎟ f ⎝RR1 2 ⎠ R 1 R 2 R1 is first surface; R2 is second. R is positive if center of curvature on transmission side. f positive if conveging lerns; negative if diverging. Lensmaker’s example n = 1.5 1 ⎛ 1 1 ⎞ =()n −1 ⎜ − ⎟ f ⎝RR1 2 ⎠ R2 = -15 cm R1 = +10 cm Incident Light 1 ⎡ 1 1 ⎤ (= 1 . 5 −⎢ 1 ) − ⎥ f ⎣10 −15⎦ f= 12 cm Light coming from ∞, focuses at focal point. R1 is first surface; R2 is second. R is positive if center of curvature on transmission side. f positive if conveging lerns; negative if diverging. Convergent Lens example • Draw some ray diagrams: Parallel ray – goes through F2 Focal ray (F1 ) – becomes parallel Central ray - undeviated h F1 s’ s f F2 f h’ '' 11 1 ' fs s h +=' s = M= − = s sf s− f s h Sign conventions for lenses (same as mirrors) • IF object & incoming ray on same side, positive s, otherwise negative. • IF image & outgoing ray on same side, positive s′, otherwise negative. • If converging lens, then f is positive; for diverging lens, f negative. '' 11 1 ' fs s h +=' s = M= − = ss f s− f s h 13. A parallel laser beam of width w1 is incident on a two lens system as shown below. w2 Each lens is converging. The second lens has a larger focal length than the first (f2> f1). What does the beam look like when it emerges from the second lens? a) The beam is converging b) The beam is diverging c) The beam is parallel to the axis with a width < w1 d) The beam is parallel to the axis with a width = w1 e) The beam is parallel to the axis with a width > w1 6. The lens in your eye a) is always a positive (i.e., converging) lens b) is always a negative (i.e., diverging) lens c) is sometimes positive, and sometimes negative, depending on whether you are looking at an object near or far away d) is positive if you are near-sighted and negative if you are far- sighted 7. The image on the back of your retina is a) inverted b) noninverted 8. The image on the back of your retina is a) real b) virtual 9. The image on the back of your retina is a) enlarged b) reduced Lecture 26, ACT 1 • A lens is used to image an object on a object screen. The right half of the lens is covered. – What is the nature of the image on the screen? lens (a) left half of image disappears (b) right half of image disappears screen (c) entire image reduced in intensity Lecture 26, ACT 1 • A lens is used to image an object on a object screen. The right half of the lens is covered. – What is the nature of the image on the screen? lens (a) left half of image disappears (b) right half of image disappears screen (c) entire image reduced in intensity • All rays from the object are brought to a focus at the screen by the lens. • The covering simply blocks half of the rays. • Therefore the intensity is reduced but the image is of the entire object! •( one more thing… In our examples of image formation, we only needed the top half of the lens to form the image) h s’ s f h’ Divergent lens example • Because the lens is divergent, f is negative: h h’ F2 F1 s f f s’ Parallel ray – goes through F2 Focal ray (F1 ) – becomes parallel Central ray - undeviated '' 11 1 ' fs s h +=' s = M= − = s sf s− f s h f is negative. Sign conventions for lenses (same as mirrors) • IF object & incoming ray on same side, positive s, otherwise negative. • IF image & outgoing ray on same side, positive s′, otherwise negative. • If converging lens, then f is positive; for diverging lens, f negative. '' 11 1 ' fs s h +=' s = M= − = ss f s− f s h Graphical images of Thin Lens Compound Lens Example Image of 1st lens becomes with ray tracing object of 2nd lens Resulting image is upright Work on board More generally…Lensmaker’s Formula 1 ⎛1 1⎞ =(n − 1⎜ ) − ⎟ f ⎝RR1 2⎠ Two arbitrary indices ⎛ ⎞⎛ ⎞ of refraction 1 n2− n 1 1 1 = ⎜ ⎟⎜ − ⎟ f ⎝ n1 ⎠⎝ R1 2 R⎠ The complete generalized case… Note: for one surface Planar, R = ∞ R > 0 if center on transmission side R < 0 otherwise Lecture 26, ACT 3 • What happens to the focal length of a lens air air when it is submerged in water? fair – In air, the focal length of a glass lens (n=1.5) is determined to be fair. When the lens is submersed in water(n=1.33), its focal length is measured to be f . What water water water fwater is the relation between fair and fwater? (a) fwater < fair (b) fwater = fair (c) fwater > fair Lecture 26, ACT 3 • What happens to the focal length of a lens air air when it is submerged in water? fair – In air, the focal length of a glass lens (n=1.5) is determined to be fair. When the lens is submersed in water(n=1.33), its water water focal length is measured to be fwater. fwater What is the relation between fair and fwater? (a) fwater < fair (b) fwater = fair (c) fwater > fair • The refraction is determined by the difference in the two indices of refraction − The smaller the difference, the less the bend the longer the focal length • Think of the case where it is glass “in glass” ! f → infinity ! • Quantitatively, look at the lensmaker’s formula 1111⎛⎞nnglass ⎛⎞glass ≈−<≈−⎜⎟11 ⎜⎟ fwater ⎝⎠nRfnRwater air ⎝⎠air 12. If you are swimming underwater (without goggles), the focusing power of the lens in your eye will a) increase b) decrease c) stay the same and sometimes inverted. Note: The “power” of a lens is defined as 1/f. The unit “diopter” is used for the power of a 1-m focal length lens. (E.g., the eyeglass lens of a slightly near-sighted person might have a power of -0.5 diopter Æ f = -2 m) and sometimes inverted. Question: How many diopters is a typical eye? Answer: f ~ 1 inch ~ 2.5 cm = 0.025 m Æ Power = 40 D Summary of Lenses and Mirrors • We have derived, in the paraxial (and thin lens) approximation, the same equations for mirrors and lenses: 1 1 1 s ′ + = M = − s s′ f s when the following sign conventions are used: Principal rays “connect” object and image Variable Mirror Lens one goes through the f > 0 concave converging center of the lens or f < 0 convex diverging mirror the other goes parallel to s > 0 real (front) real (front) the optic axis and then is s < 0 virtual (back) virtual (back) refracted or reflected through a focal point s’> 0 real (front) real (back) s’< 0 virtual (back) virtual (front) the third one is like the second one….
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