Thin Lenses
Light in air strikes a thin lens made of plastic (n=1.6). The radius of curvature of the front surface is +0.05m, the radius of the back surface is +0.1m
What are the surface powers?
Front surface: D = (1.6-1)/+ 0.05 = +12D
Back surface: D = (1-1.6)/+ 0.1 = -6D
Total power of lens is +12 -6 = +6D
*Dtotal = DFront + DBack
A real object sits 40cm in front of a PLANE mirror. What is the power of the mirror? Determine the location of the image. Is the image real or virtual?
D = 0
U + D = V > -40cm + 0 = V (little v = -40cm)
Image is virtual
*Remember object variables are the same (real obj (-u) and virtual object (+u) BUT the image variables are opposite for mirrors (real image (+v) and virtual image (-v))
Pt has myopia and astigmatism: -5.25-1.25 x 090
Draw the optical cross the spherical lens (myopia), cylindrical (astigmatism) lens and the sphero-cylindrical lens
Answer on slide 45 on Cyl lecture
T or F
A convex MIRROR has converges light (i.e. has added plus power)
F
It's the opposite of lenses
A convex mirror has diverging light (i.e. added minus power)
Slides 33-36 of Reflection lecture
T or F:
D = (n2 - n1)/r is equivalent to saying D = (nfrom - ninto)/r
F
It's equivalent to D = (ninto - nform)/r
Take a +15D thin lens in air made of high index plastic where (n=1.6).
Calculate and plot F1, F2, and N
If there is a virtual object 10cm from the lens where would the image be (describe if it's real/virtual, minified/magnified, upright/inverted)
f2 = +n2/D > +1/+15 = +6.7cm
N = where the lens in (no math)
u = +10cm
-Real image, minified and upright (check with math)
U + D = V > n1/u + D = n2/V > 1/+.10 + 15 = 1/v > v= +0.04 (real image)
M =U/V > +10/+25 = +0.4x (minified and upright)
Ray trace slide 47 on ppt
Light from a real object at 67cm strike a +4D concave mirror.
What type of image will be formed and where? Draw it out without the ray traces
Real object, so u = -0.67m > U = 1/-0.67m - -1.50
U + D = V
-1.50 +4 = +2.50
v= 1/+2.50 = +0.40m (so a real image will be formed to the left of the mirror)
M = U/D = -1.50/+2.50 = -0.60 (inverted and minified)
*concave = converging light (for a mirror)
Picture on slide 44 on Reflection lecture
Write the minus, plus and cross-cylinder form of an optical cross with +2.00 along the 090 meridian and -1.00 along the 180 meridian
Answers on slide 49-56 on Cyl lecture
Minus cyl: +2.00 - 3.00 x 090
Plus cyl: -1.00 + 3.00 x 180
Cross cyl: +2.00 x 180/ -1.00 x 090
Focal points regarding a MIRROR
Fill in the blank
1. -f1 means it will be on the ______ side
2. +f1 means it will be on the _____ side
3. -f2 means it will be on the ______ side
4. +f2 means it will be on the _____ side
1. Left
2. Right
3. Right
4. Left
*f1 and f2 have opposite signs but they are ALWAYS at the same location
-left of a concave mirror and right of a convex mirror
Slide 48 of Reflection lecture
Why is lensometer more accurate in measuring the total lens powers vs a lens clock?
Lens clock measures the from and back surface of a lens so it does not take into account the THICKNESS of a lens, but a lensometer does (true lens power is measured and is aka the vertex power)
A thin lens with meniscus shape has a front radius of r1 = +10cm and a back radius of r2 = +25cm. The lens is made of index 1.5 and is located in air.
Calculate the power, determine the cardinal points, and ray trace for the image if a real object is at 56cm and for a real object at 20cm. Describe the image also.
Answer on PPS 3 question 1 part ii and iii
A real object 40cm from a -4D convex mirror produces an image.
Where is that image and what type is it?
Real object, so u = -0.4m
U = 1/-0.4 = -2.50
U + D = V > -2.50 -4 = -6.50
v = 1/-6.50 = -0.15m
Virtual image will be formed to the right of the mirror
M = U/V = -2.50/-6.50 = +0.385 (upright and minified)
*concex = diverging light (for a mirror)
Picture on slide 46 on Reflection lecture
Given the Pt's Rx of +7.50-6.50 x 090
What is the power along the 60th meridian?
-make optical cross first
Answer on slides 64-66 on Cyl lecture
D60 = Dsph + Dcylsin2theta
Theta is the angle b/t meridian of interest and cylinder axis: 90-60 = 30
D60 = +7.50 + (-6.50)sin2(30)
D60 = +5.88
Fill in the blank:
1. 1st Purkinje-Sansom (PS) image is ___ & ___
2. 2nd PS image is ____ & ____
3. 3rd PS image is ____ & ____
4. 4th PS image is ____ & ____
1. Virtual and upright
2. Virtual and upright
3. Virtual and upright
4. Real and inverted
Slides 83-83 of Reflection lecture
A flat REFRACTIVE surface has how much power?
D = 0
B/c it does not change the vergence of light
Thin lens, meniscus shape, front radius r1 = +25cm and back radius r2 = +10cm. The lens is made of index 1.5 and is located in air.
Calculate the power, determine the cardinal points, ray trace for an image if a real object is at 35cm and if a virtual object is at 20cm. Describe the image.
Answer on PPS 3 question 2 ii and iv
A real object is 40cm from a convex mirror of radius 33.3cm immersed in water.
Do a ray trace and describe the image
Answer on slide 64 of Reflection lecture
D = -2n/r > -2(1.33)/0.333m = -8D
f1 = -n/D = -(1.33)/-8 = +16.7cm
f2 = +n/D = +1.33/-8 = -16.7cm
U + D = V same as n/u + D = n/v
> 1.33/-0.4 -8 = 1.33/v > v = -11.8cm (virtual image on the right)
M = U/V = -3.33/-11/3 = +0.29 (upright and minified)
A lens has the following Rx: +6.50-3.50 x 070
Where does it read +4.25?
Draw optical cross first
Answer on slides 67-69 on Cyl lecture
+4.25 = +6.50 + (-3.50)sin2theta
Theta = 53.3 degrees
Means +4.25 is 53.3 degrees away from either side of the 070th meridian, so at axis 16.7 and at axis 123.3
Fill in the blank:
A vertical meridian will produce a _____
line
A horizontal meridian will produce a _____ line
-Horizontal
-Vertical
Formula to calculate reflective power of a mirror
D = -2n/r
Slide 38 and 49-51 for an example
Draw all the plus and minus lens shapes and give a definition for each
Answers on slides 24-25 and 27
*Both plus and minus meniscus lenses have plus on the from and minus on the back; the total lens power depends on which outweighs the other (look at slide 26 for examples)
Relative size (largest to smallest) of the PS images for fixating at distance is:
Relative size (largest to smallest) of the PS images for fixating and accommodating at near
Distance: III > I > II > IV
Near: I > II> III > IV
The following cylinders are combined with each other:
+0.25 x 115
-1.00 x 025
Combine the cylinders and write the Rx in both plus and minus cylinder forms
Answer on slides 83-84 on Cyl lecture
Minus: +0.25 -1.25 x 025
Plus: -1.00 + 1.25 x 115
Define Euler's Law
Any 2 meridians 90 degrees apart will add up to a constant
Ex. +2.75 - 1.50 x 090
D180 + D090 = +2.75 + 0.75 = +3.50
Any 2 meridians 90 degrees apart will add up to +3.50 for this Rx
D045 + D135 = 1.75 + 1.75 = +3.50
Slide 71 on Cyl lecture
On toric surfaces:
The steeper meridian is the (power/axis) meridian
The flatter meridian is the (power/axis) meridian
Power
-has the most power/shortest radius
Axis
-has the least power/longest radius