4 Geometric Optics Mirrors and Thin Lenses Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB We have already learned the basics of Reflection and Refraction Reflection angle of incidence angle of reflection ID: 493654
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Slide1
Physics 4
Geometric OpticsMirrors and Thin Lenses
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide2
We have already learned the basics of Reflection and Refraction.
Reflection - angle of incidence = angle of reflection
Refraction - light bends toward the normal according to Snell’s Law
Now we apply those concepts to some simple types of mirrors and lenses.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide3
We have already learned the basics of Reflection and Refraction:
Reflection - angle of incidence = angle of reflection
Refraction - light bends toward the normal according to Snell’s Law
Now we apply those concepts to some simple types of mirrors and lenses.
Flat Mirror
This is the simplest mirror – a flat reflecting surface. The light rays bounce off and you see an image that seems to be behind the mirror. This is called a VIRTUAL IMAGE because the light rays do not actually travel behind the mirror. The image will appear reversed, but will be the same size and the same distance from the mirror. A typical light ray entering the eye of the viewer is shown.
The object distance is labeled do and the image distance is labeled di.
Virtual Image
d
o
d
i
Real Object
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide4
Spherical Mirrors
For curved mirrors we will assume that the shape is spherical (think of a big shiny ball, and slice off any piece of that – there’s your spherical mirror). This will make our math relatively simple, with only a couple of formulas. The hard part will be to get the negative signs correct.
The radius of curvature describes the shape of the mirror. This is the same as the radius of the big shiny ball that the mirror was cut from.
We will have two types of mirrors, depending on which direction they curve:
CONCAVE
mirrors curve toward you, and have
POSITIVE R (like the inside of the sphere).CONVEX mirrors curve away from you, and have NEGATIVE R (think of the outside of the ball).There is a point called the FOCAL POINT which is halfway between the mirror and the center.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide5
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide6
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Before we answer this let’s look at a few basic formulas for spherical mirrors.
1) The focal length is half the radius.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Remember the sign convention – if the mirror is concave R is positive. If convex, R is negative.Slide7
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Before we answer this let’s look at a few basic formulas for spherical mirrors.
1) The focal length is half the radius.
2) This formula relates the object (d
o
) and image (d
i
) positions to the focal length (f) of the mirror.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Here d
o
is always positive for mirrors, and d
i
is positive if the image is on the same side as the object (a REAL image).
To remember this, just follow the light – a real (positive) image will have light rays passing through it.
Remember the sign convention – if the mirror is concave R is positive. If convex, R is negative.Slide8
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Before we answer this let’s look at a few basic formulas for spherical mirrors.
1) The focal length is half the radius.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Remember the sign convention – if the mirror is concave R is positive. If convex, R is negative.
3) The magnification (m) of the image is related to the relative positions of the object and image.
Don’t forget the negative sign in this formula. The sign of m tells you if the image is upright (+) or inverted (-)
2) This formula relates the object (d
o
) and image (d
i
) positions to the focal length (f) of the mirror.
Here d
o
is always positive for mirrors, and d
i
is positive if the image is on the same side as the object (a REAL image).
To remember this, just follow the light – a real (positive) image will have light rays passing through it.Slide9
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
object distance
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide10
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
object distance
Now we can use formula 2 to locate the image (d
i
)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide11
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
object distance
Now we can use formula 2 to locate the image (d
i
)
This means the image will be located 1m BEHIND the mirror.
This is a VIRTUAL image.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide12
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
object distance
Now we can use formula 2 to locate the image (d
i
)
This means the image will be located 1m BEHIND the mirror.
This is a VIRTUAL image.
For the magnification, just use formula 3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide13
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
object distance
Now we can use formula 2 to locate the image (d
i
)
This means the image will be located 1m BEHIND the mirror.
This is a VIRTUAL image.
For the magnification, just use formula 3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
So the image is upright (+) and 5 times as large as the object.
We could also draw the ray diagram…Slide14
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Object
1
3
2
Image
f
d
o
d
i
Notice the 3 rays in the diagram. They all start at the object and go toward the mirror. Ray 1 through the center is easy to draw. So is ray 2, which starts out flat, then bounces off the mirror and goes through the focal point (f).
Ray 3 is the tricky one. Since the object is inside the focal point (closer to the mirror, or do<f) we can’t draw the ray through the focal point. Instead we pretend the ray came from the focal point and passed through the object on its way to the mirror, then bounced off flat.
The outgoing rays do not intersect! So we have to trace them backwards to find their intersection point behind the mirror. This is what your brain does for you every time you look in a mirror. The virtual image appears at the point where the outgoing light rays seem to be coming from.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide15
C
Convex Mirrors
These will work the same way as concave, but R and f are negative. Take a look at where the center of the sphere is – it is behind the mirror. There are no light rays there. This is why the radius is negative. Because the light rays do not go there.
The 3 typical light rays are shown.
Ray 1 points toward the center and bounces straight back.
Ray 2 starts flat and bounces off as if it is coming from the focal point.
Ray 3 starts toward the focal point and bounces off flat.
Convex Mirror – R is negative
R
f
3
2
1
object
Image (this is a virtual image behind the mirror, so d
i
is negative)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide16
C
F
V
Light In Side
d
o
> 0 Real Object
Light Out Side
d
i
> 0 Real Image
C This Side, R > 0
d
o
< 0 Virtual Object
d
i
< 0 Virtual Image
C This Side, R < 0
Optic Axis
C – Center of Curvature
R – Radius of Curvature
F – Focal Point (Same Side as C)
V – Vertex
Equations: Paraxial Approximation
Concave Mirror Illustrated
SPHERICAL MIRROR EQUATIONS
AND SIGN CONVENTION
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide17
Light In Side
d
o
> 0 Real Object
d
i
< 0 Virtual Image
C This Side, R < 0n
a – Index of
Refraction
Light Out Side
d
o < 0 Virtual Objectdi
> 0 Real ImageC This Side, R > 0nb – Index of
Refraction
REFRACTION AT SPHERICAL INTERFACE BETWEEN TWO OPTICAL MATERIALS
Illustrated Interface Has C, Center of Curvature,
On The Light Out Side, Thus R > 0
A Flat Interface Has R =
∞
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide18
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide19
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) consider the fish to be the light source, and calculate the image position for light rays exiting the bowl.
We will be using this formula:
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?Slide20
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) consider the fish to be the light source, and calculate the image position for light rays exiting the bowl.
We will be using this formula:
Here is the given information:
This radius is negative because the center of the bowl is on the same side as the light source (the fish)
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?Slide21
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) consider the fish to be the light source, and calculate the image position for light rays exiting the bowl.
We will be using this formula:
Here is the given information:
This radius is negative because the center of the bowl is on the same side as the light source (the fish)
A negative value for S’ means the image is on the same side of the interface as the object (i.e. inside the bowl in this case). So the observer will see a virtual image of the fish at the center of the bowl.
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?Slide22
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) consider the fish to be the light source, and calculate the image position for light rays exiting the bowl.
We will be using this formula:
Here is the given information:
This radius is negative because the center of the bowl is on the same side as the light source (the fish)
A negative value for S’ means the image is on the same side of the interface as the object (i.e. inside the bowl in this case). So the observer will see a virtual image of the fish at the center of the bowl.
Magnification can be found from this formula:
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?Slide23
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) consider the fish to be the light source, and calculate the image position for light rays exiting the bowl.
We will be using this formula:
Here is the given information:
This radius is negative because the center of the bowl is on the same side as the light source (the fish)
A negative value for S’ means the image is on the same side of the interface as the object (i.e. inside the bowl in this case). So the observer will see a virtual image of the fish at the center of the bowl.
Magnification can be found from this formula:
The fish appears larger by a factor of 1.33
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?Slide24
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) the light source is the sun, which is really far away (i.e. object distance is infinity). This is what they mean by “parallel rays from the sun”.
The focal point will be where the sun’s rays converge, so we need to find the image distance
d
i
.
sunlight
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?Slide25
sunlight
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) the light source is the sun, which is really far away (i.e. object distance is infinity). This is what they mean by “parallel rays from the sun”.
This radius is positive because the center of the bowl is on the opposite side as the light source (the sun)
Our given information becomes:
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
The focal point will be where the sun’s rays converge, so we need to find the image distance
d
i
. Slide26
sunlight
Image#1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) the light source is the sun, which is really far away (i.e. object distance is infinity). This is what they mean by “parallel rays from the sun”.
This radius is positive because the center of the bowl is on the opposite side as the light source (the sun)
This
woul
d be the focal point if the light stayed in the water until it got there. Since the light rays travel back into the air at the other side of the bowl, we will have to do a second calculation to find the final position of the image.
The first image becomes the object for the interface at the right side of the bowl.
Our given information becomes:
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
The focal point will be where the sun’s rays converge, so we need to find the image distance
d
i
. Slide27
sunlight
Image #2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
The first calculation gave
us an image position to the right of the bowl, a distance of 28cm to the right.
Thus the new object distance will be d
o
=-28cm.
This radius is
negative
because the center of the bowl is on
incoming light side (inside the bowl)
This
final image is 14cm to the right of the bowl, and is the focal point of the “lens” formed by the bowl of water. Sinc
e this focal point is outside the bowl, our fish is safe.
Problem
34.17
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
The values we know for the 2
nd
calculation are:Slide28
Two Basic Types of Lenses
CONVERGING
f is positive
Thicker in middle
Object outside focal point = real image
Object inside focal point = virtual image
Focal Point
DIVERGING
f is negative
Thinner in middle
Real object always gives a virtual image
Focal PointSlide29
Ray Diagrams for Lenses
There are 3 convenient rays to draw for a thin lens. The profile of the lens is just shown for reference – all the bending of the light occurs at the plane through the center of the lens.Slide30
Surface 1
Surface 2
Light In Side
d
o
> 0 Real Object
d
i
< 0 Virtual Image
C
1
This Side, R
1
< 0
C
2
This Side, R
2
< 0
Light Out Side
do
< 0 Virtual Objectdi > 0 Real ImageC1
This Side, R1 > 0C2 This Side, R
2 > 0
n – Index of Refraction
C
1
– Center of Curvature, Surface 1C2 – Center of Curvature, Surface 2
Illustrated Lens is Double Convex Converging
With C
1 on the Light Out Side and C2 on the Light In Side
Equations:
THIN LENS EQUATIONS
AND SIGN CONVENTION
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide31
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Radius=15cm
Radius=20cm
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide32
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Radius=15cm
Radius=20cm
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
To find the focal length we use the thin lens equation:Slide33
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
R
2
=+15cm
R
1
=+20cm
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
To find the focal length we use the thin lens equation:
Light traveling this direction
The difficult part is to get the signs correct for the radii. We can suppose the light is coming from the left, so the light encounters the 20cm side first. Since the center of that 20cm-radius circle is on the other side (where the light rays are going to end up) we call this radius positive – so
R
1
=+20cm
.
Similarly, the 15cm-radius circle has its center on the other side, so this is also positive:
R
2
=+15cm
Your basic rule of thumb is this: follow the light rays – they end up on the positive side.Slide34
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
R
2
=+15cm
R
1
=+20cm
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
To find the focal length we use the thin lens equation:
Light traveling this direction
For extra bonus fun, try calculating the focal length when the light comes from the other side – so the 15cm radius is encountered first.
You ought to get the same answer for the focal length.Slide35
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) we can use the formula:
f=-100cm
d
o
=+
50cmSlide36
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) we can use the formula:
f=-100cm
d
i
=-
33.3cm
d
o
=+
50cmSlide37
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) we can use the formula:
f=-100cm
The height of the image comes from our magnification formula:
The image is virtual, upright, and 8cm tall.
d
i
=-
33.3cm
d
o
=+
50cmSlide38
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) we can use the formula:
f=-100cm
The height of the image comes from our magnification formula:
The image is virtual, upright, and 8cm tall.
d
i
=-
33.3cm
d
o
=+
50cmSlide39
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) we can use the formula:
f=-100cm
The height of the image comes from our magnification formula:
The image is virtual, upright, and 8cm tall.
d
i
=-
33.3cm
d
o
=+
50cmSlide40
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is this image real or virtual? Draw the ray diagram.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part b) we can use the formula:
f=-100cm
The height of the image comes from our magnification formula:
The image is virtual, upright, and 8cm tall.
The red ray in our diagram is initially headed for the focal point on the other side of the lens at x=+100cm.
The lens deflects it parallel to the axis, and we trace it back to find the image (at the intersection with the other 2 rays)
d
i
=-
33.3cm
d
o
=+
50cm