Thin Lenses 1/p + 1/q = 1/f 1/f = (n -1) (1/R1 - 1/R2) R is > 0 if convex toward the object and R<0 if concave toward the object If we assume object is on left: R2 < 0 R1 > 0  f > 0 R2 > 0 R1 > 0 |R1| < |R2|  f > 0 R1 > 0 R2 =   f > 0 R2 > 0 R1 < 0  f < 0 R2 > 0 R1 > 0 |R1| > |R2|  f < 0 R2 > 0 R1 =   f < 0 A lens which is fatter in center than edges will have f > 0 and is called a converging lens. A lens which is thinner in center than edges will have f < 0 and is called a diverging lens.

1/f = (n-1) (1/R1 – 1/R2) 1/p + 1/q = 1/f M = h’ / h = -q/p Problem: An object 3 cm high is 20 cm from a converging lens with R1 = 5 cm, R2 = - 10 cm, and n = 3/2. Where is the image and how large is it? Is it inverted or upright? Real or virtual?

1/f = (3/2 -1) (1/5cm – (-1/10cm)) = +3/20cm f = + 6.67 cm 1/f = (n-1) (1/R1 – 1/R2) 1/p + 1/q = 1/f M = h’ / h = -q/p Problem: An object 3 cm high is 20 cm from a converging lens with R1 = 5 cm, R2 = - 10 cm, and n = 3/2. Where is the image and how large is it? Is it inverted or upright? Real or virtual? 1/f = (3/2 -1) (1/5cm – (-1/10cm)) = +3/20cm f = + 6.67 cm p = 20 cm: 1/q = 1/f – 1/p = 3/20cm – 1/20cm = 1/10cm q = + 10 cm: real image h’ = -(q/p)h = - (10/20) (3 cm) = -1.5 cm (inverted) not to scale

1/f = (n-1) (1/R1 – 1/R2) 1/p + 1/q = 1/f M = h’ / h = -q/p Problem: An object 3 cm high is 2 cm from a converging lens with R1 = 5 cm, R2 = - 10 cm, and n = 3/2. Where is the image and how large is it? Is it inverted or upright? Real or virtual?

1/f = (3/2 -1) (1/5cm – (-1/10cm)) = +3/20cm f = + 6.67 cm 1/f = (n-1) (1/R1 – 1/R2) 1/p + 1/q = 1/f M = h’ / h = -q/p Problem: An object 3 cm high is 2 cm from a converging lens with R1 = 5 cm, R2 = - 10 cm, and n = 3/2. Where is the image and how large is it? Is it inverted or upright? Real or virtual? 1/f = (3/2 -1) (1/5cm – (-1/10cm)) = +3/20cm f = + 6.67 cm p = 2 cm: 1/q = 1/f – 1/p = 3/20cm – 1/2cm = -7/20cm q = - 2.86 cm: virtual image h’ = -(q/p)h = - (-2.86/2) (3 cm) = +4.29 cm (upright) not to scale

1/f = (n-1) (1/R1 – 1/R2) 1/p + 1/q = 1/f M = h’ / h = -q/p Problem: An object 3 cm high is 2 cm from a diverging lens with R1 = -5 cm, R2 = + 10 cm, and n = 3/2. Where is the image and how large is it? Is it inverted or upright? Real or virtual?

1/f = (3/2 -1) (-1/5cm – (1/10cm)) = -3/20cm f = - 6.67 cm 1/f = (n-1) (1/R1 – 1/R2) 1/p + 1/q = 1/f M = h’ / h = -q/p Problem: An object 3 cm high is 20 cm from a diverging lens with R1 = -5 cm, R2 = + 10 cm, and n = 3/2. Where is the image and how large is it? Is it inverted or upright? Real or virtual? 1/f = (3/2 -1) (-1/5cm – (1/10cm)) = -3/20cm f = - 6.67 cm p = 20 cm: 1/q = 1/f – 1/p = -3/20cm – 1/20cm = -1/5cm q = - 5 cm: virtual image h’ = -(q/p)h = - (-5/20) (3 cm) = +0.75 cm (upright) not to scale

Thin Lens Ray Diagrams 1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. (Make lens as high as needed.) 2) A ray through the center of the lens hits two ~ parallel surfaces, so comes out parallel (with very tiny displacement) to the way it came in: this ray continues straight through in the thin lens limit. [This is the analog of the “extra” mirror characteristic ray.] 3) An incoming ray parallel to the axis will refract through the focal point a) on back side of lens for a converging lens; b) as if came from the focal point on front of lens for diverging lens. 4) A ray will come out parallel to the principal axis if it a) passed through or came from the focal point on the front side for a converging lens; b) was headed toward the focal point on the back of a diverging lens. 5) If needed, extend the rays to the front side of the lens (from where they appear to originate) for a virtual image. [Note: no analog of mirror characteristic ray through center of curvature.]

Converging lens with p >f 1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens hits two ~ parallel surfaces, so comes out parallel to the way it came in: this ray continues straight through in the thin lens limit. [Since it is going through two ~ parallel surfaces, it will be displaced, but if the lens is thin, the displacement  0.] h f p f

1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. 3) An incoming ray parallel to the axis will refract through the focal point on back side of lens for a converging lens. h f p f

1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. 3) An incoming ray parallel to the axis will refract through the focal point on back side of lens for a converging lens. 4) A ray will come out parallel to the principal axis if it passed through or came from the focal point on the front side for a converging lens. h f q p f h’ 1/q = 1/f – 1/p; h’ = h(-q/p) If p > f, real inverted image (e.g. projector: if p~f  q and h’ are large)

Converging Lens with p < f 1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. h f p f

1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. 3) An incoming ray parallel to the axis will refract through the focal point on back side of lens for a converging lens. h f p f

1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. 3) An incoming ray parallel to the axis will refract through the focal point on back side of lens for a converging lens; 4) A ray will come out parallel to the principal axis if it passed through or came from the focal point on the front side for a converging lens. (Make lens as high as needed.) h’ 5) If needed, extend the rays to the front side of the lens (where they look like they come from) for a virtual image. h -q f p f 1/q = 1/f – 1/p; h’ = h(-q/p) If p < f, virtual, upright image (e.g. magnifying glass)

Diverging Lens 1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. h f p f

1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. 3) An incoming ray parallel to the axis will refract through the focal point as if came from the focal point on front of lens for diverging lens. h f p f

1) Draw accurate scale diagrams, with object (size h, distance p from lens) and with focal points on both sides of lens. 2) A ray through the center of the lens continues straight through. 3) An incoming ray parallel to the axis will refract through the focal point as if came from the focal point on front of lens for diverging lens. 4) A ray will come out parallel to the principal axis if it was headed toward the focal point on the back of a diverging lens. 5) If needed, extend the rays to the front side of the lens (where they look like they come from) for a virtual image. h f h’ -q p f 1/q = 1/f – 1/p; h’ = h(-q/p) For diverging lens, always virtual, upright, demagnified image

Problem: An object, h = 3 cm high, is p1 = 20 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 8 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? Note that this figure is not to scale! Since light from object first goes through left lens before going through right lens, treat the lenses one at a time: the image of the left lens acts as the object for the right lens.

1/q1 = 1/f1-1/p1 = 1/5 cm – 1/20cm = 3/20cm q1 = +6.67 cm (real image) Problem: An object, h = 3 cm high, is p1 = 20 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 8 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? h1’ h1’ q1 p2 1/q1 = 1/f1-1/p1 = 1/5 cm – 1/20cm = 3/20cm q1 = +6.67 cm (real image) h1’ = -(q1/p1)h = -(6.67/20)(3 cm) = -1 cm p2 = L – q1= (20-6.67) cm = + 13.33 cm Note that this figure is not to scale!

h2’ = -(q2/p2) h1’ = -(20/13.33)(-1cm) h2’ = + 1.5 cm (upright) Problem: An object, h = 3 cm high, is p1 = 20 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 8 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? h2’ h2 = h1’ p2 1/q2 = 1/f2 – 1/p2 1/q2 = 1/8cm – 1/13.33cm = .050/cm q2 = + 20 cm (real image) h2 = h1’ = -1 cm h2’ = -(q2/p2) h1’ = -(20/13.33)(-1cm) h2’ = + 1.5 cm (upright) 1/q1 = 1/f1-1/p1 = 1/5 cm – 1/20cm = 3/20cm q1 = +6.67 cm (real image) h1’ = -(q1/p1)h = -(6.67/20)(3 cm) = -1 cm p2 = L – q1= (20-6.67) cm = + 13.33 cm Note that this figure is not to scale!

Problem: An object, h = 3 cm high, is p1 = 7 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 8 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? Since p1 is smaller than the previous problem, expect q1 to be larger.

Problem: An object, h = 3 cm high, is p1 = 7 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 8 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? q1 p2 h2 = h1’ = -7.5 cm p2=L-q1= 2.5 cm h1’ = h2 [Figure not to scale] 1/q1 = 1/f1-1/p1 1/q1 = 1/5cm-1/7cm q1 = 17.5 cm h1’ = -(q1/p1)h1 = -7.5 cm

Problem: An object, h = 3 cm high, is p1 = 7 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 8 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? q2 q1 p2 h2 = h1’ = -7.5 cm p2=L-q1= 2.5 cm h1’ = h2 1/q1 = 1/f1-1/p1 1/q1 = 1/5cm-1/7cm q1 = 17.5 cm h1’ = -(q1/p1)h1 = -7.5 cm 1/q2 = 1/f2-1/p2 1/q2 = 1/8cm – ½.5cm q2 = -3.64cm (virtual) h2’ = -(q2/p2)h2 = - 10.9 cm (inverted) h2’ [Figure not to scale]

Problem: An object, h = 3 cm high, is p1 = 6 Problem: An object, h = 3 cm high, is p1 = 6.5 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 10 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? Since p1 is smaller than the previous problem, expect q1 to be larger. [Also changed f2.]

Problem: An object, h = 3 cm high, is p1 = 6 Problem: An object, h = 3 cm high, is p1 = 6.5 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 10 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? h1’ 1/q1 = 1/f1-1/p1 1/q1 = 1/5cm – 1/6.5cm q1 = 21.67 cm h1’ = (-q1/p1)h = -10 cm q1

Problem: An object, h = 3 cm high, is p1 = 6 Problem: An object, h = 3 cm high, is p1 = 6.5 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 10 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? -p2 Note that these rays will actually never reach q1 because the second lens will change their direction (if large diameter): h1’ acts as a virtual object to the second lens, with p2 = -1.67 cm. 1/q1 = 1/f1-1/p1 1/q1 = 1/5cm – 1/6.5cm q1 = 21.67 cm h1’ = (-q1/p1)h = -10 cm h1’ q1 Virtual Object: Rays were heading toward that point (if the lens weren’t present), but its position is behind the lens, so acts as an object for the lens with p < 0.

Problem: An object, h = 3 cm high, is p1 = 6 Problem: An object, h = 3 cm high, is p1 = 6.5 cm to the left of a converging lens with focal length f1 = 5 cm. A second converging lens, with focal length f2 = 10 cm, is L = 20 cm to the right of the first lens. Where is the image formed by the second lens, how large is it, and is it real or virtual and inverted or upright? -p2 q2 h2 = h1’ = -10 cm p2 = -1.67 cm 1/q2 = 1/f2 – 1/p2 1/q2 = 1/10cm-(-1/1.67)cm q2 = +1.43 cm: real image (light really goes there) h2’ = -(q2/p2)h1’ h2’ = -(1.43)(-10cm)/(-1.67) h2’ = - 8.60 cm (inverted!) h2’ h2