mirrors and lenses

mirrors and lenses PHY232 – Spring 2007 Jon Pumplin http://www.pa.msu.edu/~pumplin/PHY232 (Ppt courtesy of Remco Zegers)

we saw… • that light can be reflected or refracted at boundaries between material with a different index of refraction. • by shaping the surfaces of the boundaries we can make devices that can focus or otherwise alter an image. • Here we focus on mirrors and lenses for which the properties can be described well by a few equations. PHY232 - Pumplin - Mirrors and lenses

the plane (=flat) mirror p -q • in the previous chapter we already studied flat mirrors. • Distance from the object to the mirror is the object distance p • Distance from the image to the mirror is the image distance -q • in case of a flat mirror, an observer sees a virtual image, meaning that the rays do not actually come from it. • the image size (h’ ) is the same as the object size (h), meaning that the magnification h’/h=1 • the image is not inverted NOTE: a virtual image cannot be projected on a screen but is ‘visible’ by the eye or another optical instrument. PHY232 - Pumplin - Mirrors and lenses

object image question • You are standing in front (say 1 m) of a mirror that is less high than your height. Is there a chance that you can still see your complete image? • a) yes b) no PHY232 - Pumplin - Mirrors and lenses

ray diagrams • to understand the properties of optical elements we use ray diagrams, in which we draw the most important elements and parameters to understand the elements h h’ p -q PHY232 - Pumplin - Mirrors and lenses

a light ray passing through the center of curvature will be reflected back upon itself because it strikes the mirror normally to the surface. a light ray traveling parallel to the central axis of the mirror will be reflected to the focal point F, with FM=CM/2 The distance FM is called the focal length f. concave mirrors F M C C: center of mirror curvature F: focal point PHY232 - Pumplin - Mirrors and lenses

I step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F). step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis) the image of the top of the object is located where the reflected rays meet construct the image I concave mirrors: an object outside F O F step 3: note that a ray from the bottom of the object just reflects back. PHY232 - Pumplin - Mirrors and lenses

concave mirrors: an object outside F O I F • The image is: • inverted (upside down) • real (light rays pass through it) • For this location, image is smaller than object • If interchange image and object, image would • be bigger than object. PHY232 - Pumplin - Mirrors and lenses

concave mirrors: object outside F O I F distance object-mirror: p distance image-mirror: q distance focal point-mirror: f • mirror equation: 1/p + 1/q = 1/f • given p,f this equation can be used to calculate q • magnification: M = -q/p • can be used to calculate magnification. • if negative: the image is inverted • if smaller than 1, object is demagnified PHY232 - Pumplin - Mirrors and lenses

example • An object is placed 12 cm in front of a a concave mirror with focal length 5 cm. What are: • a) the location of the image • b) the magnification • 1/p + 1/q = 1/f so 1/12 + 1/q = 1/5 • 1/q = 1/5 - 1/12 so q = 8.57 cm • b) M = -q/p = -8.57/12 = -0.71 • this means that size of the image is only 71% of the • the size of the object and that it is inverted. • c) Image is real (whenever q>0, I.e., whenever M > 0) PHY232 - Pumplin - Mirrors and lenses

step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F). step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis) the image of the top of the object is located where the reflected rays meet: in this you must draw virtual rays on the other side of the lens creates a magnified image (“shaving mirror”) concave mirrors: an object inside F F O I • the image is: • not inverted • virtual • magnified step 3: note that a ray from the bottom of the object just reflects back. PHY232 - Pumplin - Mirrors and lenses

concave mirrors: an object inside F F O I • the image is: • not inverted • virtual • magnified The lens equation and equation for magnification are still valid. However, since the image is now on the other side of the mirror, its sign should be negative PHY232 - Pumplin - Mirrors and lenses

example • an object is placed 2 cm in front of a lens with a focal length of 5 cm. What are the a) image distance and b) the magnification? • 1/p + 1/q = 1/f so 1/2 + 1/q = 1/5 • 1/q = 1/5 - 1/2 so q = -3.3 cm (note the ‘-’ sign!) • b) M = -q/p = -(-3.3)/2 = +1.65 • this means that size of the image is 65% larger than • the size of the object and that it is not inverted (+). PHY232 - Pumplin - Mirrors and lenses

demo: the virtual pig PHY232 - Pumplin - Mirrors and lenses

I step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F). step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis) the image of the top of the object is located where the reflected rays meet construct the image I (examples: garden ball, rearview mirrors that warn “objects are closer than they appear”) convex mirrors: (p>|f|or p<|f| doesn’t matter) O F F is now located on the other side of the mirror step 3: note that a ray from the bottom of the object just reflects back. PHY232 - Pumplin - Mirrors and lenses

convex mirrors O I F F is now located on the other side of the mirror • the image is: • not inverted • virtual • demagnified The lens/mirror equation and equation for magnification are still valid. However, since the image and focal point are now on the other side of the mirror, their signs should be negative PHY232 - Pumplin - Mirrors and lenses

example • an object with a height of 3 cm is placed 6 cm in front of a convex mirror, with f=-3 cm. What are a) the image distance and b) the magnification? answer a): 1/p+1/q=1/f with p=6 cm f=-3 cm 1/6 + 1/q = -1/3 so q=-2 cm b) M=-q/p=-(-2)/6=1/3 the image is only 33% of the height of the object. PHY232 - Pumplin - Mirrors and lenses

Mirrors: an overview • mirror equation 1/p + 1/q = 1/f • F = R/2 where R is the radius of the mirror • magnification: M = -q/p PHY232 - Pumplin - Mirrors and lenses

Lenses • Lenses function by refracting light at their surfaces • Their action depends on • radii of the curvatures of both surfaces • the refractive index of the lens • converging (positive lenses) have positive focal length and are always thickest in the center • diverging (negative lenses) have negative focal length and are thickest at the edges + used in drawings - PHY232 - Pumplin - Mirrors and lenses

lensmakers equation object 1 2 R2 f: focal length of lens n: refractive index of lens R1 radius of front surface R2 radius of back surface R1 R2 is negative if the center of the circle is on the left of curvature 2 of the lens R1 is positive if the center of the circle is on the right of curvature 1 of the lens if the lens is not in air then (nlens-nmedium) PHY232 - Pumplin - Mirrors and lenses

example object • Given R1=10 cm and R2=5 cm, what is the focal length? The lens is made of glass (n=1.5) 1 2 R2 R1 R1 is on the right of the curvature, so positive +10 cm R2 is on the left of the curvature, so negative –5 cm n=1.5 1/f=0.5(0.1-(-0.2))=0.15 f=+6.67 cm PHY232 - Pumplin - Mirrors and lenses

example 2 object • Given R1=5 cm and R2=10 cm, what is the focal length? The lens is made of glass (n=1.5) 1 2 R1 R2 R1 is on the left of curvature 1 so R1=-5 cm R2 is on the right of curvature 2 so R2=+10 cm n=1.5 1/f=0.5(-0.2-0.1)=-0.15 f=-6.67 cm PHY232 - Pumplin - Mirrors and lenses

I 1) A ray parallel to the central axis will be bend through the focal point 2) A ray through the center of the lens will continue unperturbed 3) A ray through the focal point of the lens will be bend parallel to the central axis 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). converging lens p>f O F F + A real inverted image is created. The magnification depends on p: |M| can be <1, 1 or >1 PHY232 - Pumplin - Mirrors and lenses

lens equation I O F F + The equation that connects object distance p, image distance q and focal length f is (just like for mirrors): 1/p + 1/q = 1/f Similarly for the magnification: M=-q/p q is positive if the image is on the opposite side of the lens as the object NOTE THAT THIS IS DIFFERENT THAN THE CASE FOR MIRRORS. In either case, q >0 means the image is on the side where the light from the lens or mirror is. PHY232 - Pumplin - Mirrors and lenses

example • an object is put 20 cm in front of a positive lens, with focal length of 12 cm. a) What is the image distance q? b) What is the magnification? a) P = 20 cm, f = 12 cm use 1/p + 1/q = 1/f 1/20 + 1/q = 1/12 solve for q gives q=30 cm b) M = -q/p = -30/20 = -1.5 The image is inverted (M negative). The image is magnified (|M|>1) PHY232 - Pumplin - Mirrors and lenses

I 1) A ray parallel to the central axis will be bend through the focal point 2) A ray through the center of the lens will continue unperturbed 3) A ray through the focal point of the lens will be bend parallel to the central axis 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). converging lens p<f F O F + A virtual non-inverted image is created. Magnification >1 PHY232 - Pumplin - Mirrors and lenses

example • an object is put 2 cm in front of a positive lens, with focal length of 3 cm. a) What is the image distance q? b) What is the magnification? a) p=2 cm, f=3 cm use 1/p + 1/q = 1/f 1/2 + 1/q = 1/3 solve for q gives q=-6 cm NOTE q is negative which means it is on the same side of the lens as the object b) M=-q/p=-(-6)/2=+3 The image is not inverted (M positive). The image is magnified (|M|>1) The image is virtual PHY232 - Pumplin - Mirrors and lenses

rays of light can still make it to any point on the image, but there are less of them, so less bright (only) question • An object is placed in front of a converging (positive) lens with the object distance larger than the focal distance. An image is created on a screen on the other side of the lens. Then, the lower half of the lens is covered with a piece of wood. Which of the following is true: • a) the image on the screen will become less bright only • b) half of the image on the screen will disappear only • c) half of the image will disappear and the remainder of the image will become less bright. I O F F + screen PHY232 - Pumplin - Mirrors and lenses

NOT CORRECT PHY232 - Pumplin - Mirrors and lenses

I • A ray parallel to the central axis will be bend so that the ray passes • through the focal point IN FRONT of the lens 2) A ray through the center of the lens will continue unperturbed 3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). diverging lens p>|f| O F F - A virtual non-inverted image is created. The magnification |M|<1 PHY232 - Pumplin - Mirrors and lenses

example • an object is put 5 cm in front of a negative lens, with focal length of -3 cm. a) What is the image distance q? b) What is the magnification? a) p=5 cm, f=-3 cm use 1/p + 1/q = 1/f 1/5 + 1/q = -1/3 solve for q gives q=-1.88 cm b) M=-q/p=-(-1.88)/5=+0.375 The image is non-inverted (M positive). The image is demagnified (|M|<1) PHY232 - Pumplin - Mirrors and lenses

I • A ray parallel to the central axis will be bend so that the ray passes • through the focal point IN FRONT of the lens 2) A ray through the center of the lens will continue unperturbed 3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). diverging lens p<|f| F F O - A virtual non-inverted image is created. The magnification |M|<1 similar to case with p>|f| PHY232 - Pumplin - Mirrors and lenses

example • an object is put 2 cm in front of a negative lens, with focal length of -3 cm. a) What is the image distance q? b) What is the magnification? a) p = 2 cm, f = -3 cm use 1/p + 1/q = 1/f 1/2 + 1/q = -1/3 solve for q gives q = -1.2 cm b) M = -q/p = -(-1.2)/2 = +0.6 The image is non-inverted (M positive). The image is demagnified (|M|<1) PHY232 - Pumplin - Mirrors and lenses

lenses, an overview • mirror or lens equation 1/p + 1/q = 1/f • mirror or lens magnification: M = -q/p • lens makers equation: 1/f=(n-1)(1/R1-1/R2) PHY232 - Pumplin - Mirrors and lenses

chromatic aberrations Chromatic aberrations are due to light of different wavelengths having a different index of refraction Can be corrected by combining lenses/mirrors If n varies with wavelength, the focal length f changes with wavelength PHY232 - Pumplin - Mirrors and lenses

two lenses • an object, 1 cm high, is placed 5 cm in front of a converging mirror with a focal length of 3 cm. This setup is placed in front of a diverging mirror with a focal length of –5 cm. The distance between the two lenses is 10 cm. Where is the image located, and what are its properties? + - 3cm 5cm 5 cm 15 cm PHY232 - Pumplin - Mirrors and lenses

I2 I1 5+7.5= 12.5 cm 15-1.67=13.33 cm consider the converging lens first: 1/p+1/q=1/f so 1/5+1/q=1/3 so q=7.5 cm M=-q/p=-7.5/5=-1.5, so 1.5x1cm=1.5 cm high A real inverted magnified image I1 consider the action of diverging lens on the image constructed above 1/p+1/q=1/f so 1/(2.5) + 1/q = 1/(-5) q=-1.67 cm M=-q/p=-(-1.67)/2.5=0.67, so 0.67x 1.5=1 cm high A virtual non-inverted demagnified image I2 relative to I1 answer + - O 5cm 3cm 15 cm 5 cm PHY232 - Pumplin - Mirrors and lenses

mirrors and lenses