Lenses Types of Lenses Locating images using ray

Lenses Types of Lenses Locating images using ray diagrams Calculations involving Lenses

Lenses Recall mirrors have one reflective & one opaque surface. Lenses on the other hand have no opaque surface. So instead of light being reflected as in mirrors, lights in lens will be refracted (as it goes through from air to the lens and back to air)

Basic Lens Shapes Converging lens: • Thickest in the middle • Parallel incident rays will be refracted and converge at a single point

Basic Lens Shapes Diverging lens: • Thinnest in the middle • Parallel incident rays will be refracted and diverge (spread apart)

Simplifying Path of Light Rays In reality, as light goes from AIR to LENS then out to AIR again, you see 2 refractions (bending of light). To simplify it, we only show 1 refraction through central line. Incident ray Emergent ray

Terminology of Converging Lens Different terminology than the ones used for mirrors. Optical Centre F’ Secondary Principal focus O F Principal focus Principal axis

Terminology of Diverging Lens Different terminology than the ones used for converging lens.

Converging Lens - Rules A ray parallel to principal axis. A ray through the secondary principal focus (F’) is refracted through the parallel to principal axis principal focus (F) Principal axis 2 F’ � � � F’ F = principal focus F’ = secondary principal focus O = optical centre O F 2 F A ray through Optical centre (O) continues straight through without being refracted.

Converging Lens – Real vs. Virtual Image � If Image is on the OPPOSITE SIDE of the lens from the object = REAL Image � If Image is on the SAME SIDE as object = Virtual Image This is the opposite of mirrors! Don’t get them mixed up. VIRTUAL Image Object REAL Image

Converging Lens – Finding the image �Use 2 of the rules to find the image 2 F’ F’ O F 2 F

Converging Lens Investigation � Use the templates given to investigate and complete the following table: Beyond 2 F’ Size Attitude Location Type At 2 F’ Between 2 F’ & F’ At F’ Between F’ & lens

Converging Lens -Object beyond 2 F’ When object is beyond 2 F’, the image will be: S – smaller A – inverted L – between F and 2 F T – Real 2 F’ F’ O F 2 F

Converging Lens -Object at 2 F’ When object is at 2 F’, the image will be: S – same size A – inverted L – at 2 F T – Real 2 F’ F’ O F 2 F

Converging Lens -Object between 2 F’ and F’ When object is between 2 F’ and F’, the image will be: S – larger A – inverted L – beyond 2 F T – Real 2 F’ F’ O F 2 F

Converging Lens -Object at F’ When object is at F’ NO IMAGE – lines are parallel 2 F’ F’ O F 2 F

Converging Lens -Object at F’ When object is between F’ and the lens, the image will be: S – larger A – UPRIGHT L – same side as object T – VIRTUAL 2 F’ F’ Extend your refracted ray BACKWARDS to locate the image O F 2 F

Remember: The only time you’ll get VIRTUAL image with converging lens is when the object is between F’ and O Extend your refracted ray BACKWARDS to locate virtual image 2 F’ F’ O F 2 F

Converging Lens Investigation: Results Beyond 2 F’ At 2 F’ Between 2 F’ & F’ At F’ Size Smaller Same size Larger Attitude Inverted Upright Location Between 2 F At 2 F and F Beyond 2 F Same side as object Type Real Virtual Real NO IMAGE Between F’ & lens Larger

Diverging Lens

Diverging Lens- Rules A ray parallel to principal axis A ray that APPEARS TO PASS is refracted AS IF IT HAD COME through the secondary through the principal focus (F) principal focus (F’) is refracted parallel to p. a Then EXTEND refracted ray backwards! Principal axis 2 F F O F’ 2 F’ A ray through Optical centre (O) continues straight through Note the DIFFERENCE IN PLACEMENT of F and F’ in

Diverging Lens- Finding the image Use 2 of the rules to locate the image: Principal axis 2 F F O F’ 2 F’

Diverging Lens Investigation � Use the templates given to investigate and complete the following table: Beyond 2 F Size Attitude Location Type At 2 F Between 2 F & F At F Between F & lens

Diverging Lens Investigation: Results Always the SAME image characteristics no matter where the object is located: ANY location Size Smaller Attitude Upright Location Same side as object Type Virtual

Calculations involving Lenses

1 + 1 = 1 do d i f Thin Lens Equation do di Image Object 2 F’ O F’ f F f 2 F

Thin Lens Equation 1 + 1 = 1 do d i f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens

Thin Lens Equation 1 + 1 = 1 do d i f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens Sample Problem: A converging lens has a focal length of 17 cm. A candle is located 48 cm from the lens. What type of image will be formed and where will it be located?

Thin Lens Equation 1 + 1 = 1 do d i f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens Given: f = 17 cm (converging lens so +) do = 48 cm (do always +) Required: a) type of image? b) d i ? Analysis: Rearrange thin lens equation 1 = 1 - 1 di f do

Thin Lens Equation 1 + 1 = 1 do d i f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens Solution: 1 = 1 – 1. di 17 cm 48 cm 1 = 0. 038 cm-1 di Use inverse function: di = 26 cm Since di is positive, it’s real image. Statement: The image is real image and is located 26 cm from the lens, opposite to the object

Magnification Equation M = h i = – di ho do do di ho Object Image 2 F’ O F’ f F f 2 F hi

Magnification Equation M = hi = – d i ho do h 0 and h 1 are: + when measured upward – when measured downward M is: + for upright image – for inverted image

Magnification Equation M = hi = - d i Sample Problem : A toy of height 8. 4 cm is balanced in front of converging lens. h 0 and h 1 are: + when upward – when downward An inverted, real image of height 23 cm is noticed on the other side of the lens. M is: + for upright image – for inverted image What is the magnification of the lens? ho do

Magnification Equation M = hi = - d i ho do h 0 and h 1 are: + when upward – when downward M is: + for upright image – for inverted image Given: ho = 8. 4 cm (upward, so +) hi = – 23 cm (inverted, so – ) Required: M? Analysis: Use M = hi ho

Magnification Equation M = hi = - d i ho do h 0 and h 1 are: + when upward – when downward M is: + for upright image – for inverted image Solution: M = hi = – 23 cm ho 8. 4 cm M = – 2. 7 Statement: The lens has a magnification of – 2. 7