Physics 6C

Geometric OpticsMirrors and Thin Lenses

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We have already learned the basics of Reflection and Refraction.Reflection - angle of incidence = angle of reflectionRefraction - light bends toward the normal according to Snell’s LawNow we apply those concepts to some simple types of mirrors and lenses.

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We have already learned the basics of Reflection and Refraction:Reflection - angle of incidence = angle of reflectionRefraction - light bends toward the normal according to Snell’s LawNow we apply those concepts to some simple types of mirrors and lenses.

Flat MirrorThis 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 S and the image distance is labeled S’.

Virtual Image

S S’

Real Object

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Spherical MirrorsFor 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.

Concave Mirror – R is positive Convex Mirror – R is negative

RC

RC

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Shiny sideShiny side

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Optical Axis

Focal Point

Ray 1 through the center

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Optical Axis

Focal Point

Ray 1 through the center

Ray 1 reflects directly back

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Optical Axis

Focal Point

Ray 2 through the focal point

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Optical Axis

Focal Point

Ray 2 through the focal point

Ray 2 reflects parallel to axis

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Optical AxisFocal Point

Ray 3 comes in parallel to axis

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Optical AxisFocal Point

Ray 3 reflects through focal point

Ray 3 comes in parallel to axis

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

Optical Axis

Object 1

3

2

Image

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

We will learn 2 techniques for dealing with mirrors (and lenses):• Graphical – draw the light rays and the image is at their intersection.• Formula – use a couple of formulas to locate and describe an image.

First the Graphical Method:For a spherical mirror there are 3 basic rays that you can draw:1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.3) Any ray that starts parallel to the optical axis is reflected back through the focal point.

(opposite of 2)

All 3 rays shown with the image at their intersection

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 ZacconeFor Campus Learning Assistance Services at UCSB

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.

2Rf

1) The focal length is half the radius.

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Remember the sign convention – if the mirror is concave R is positive. If convex, R is negative.

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.

2Rf

1) The focal length is half the radius.

2) This formula relates the object (S) and image (S’) positions to the focal length (f) of the mirror.

S1

S1

f1

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Here S is always positive for mirrors, and S’ 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.

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.

2Rf

1) The focal length is half the radius.

2) This formula relates the object (S) and image (S’) positions to the focal length (f) of the mirror.

S1

S1

f1

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Here S is always positive for mirrors, and S’ 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.

3) The magnification (m) of the image is related to the relative positions of the object and image.

SS

yym

Don’t forget the negative sign in this formula. The sign of m tells you if the image is upright (+) or inverted (-)

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:m2.0S

m25.0fm5.0R

focal length

object distance

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

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:m2.0S

m25.0fm5.0R

focal length

object distanceNow we can use formula 2 to locate the image (S’)

m1SS1

2.01

25.01

S1

S1

f1

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

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:m2.0S

m25.0fm5.0R

focal length

object distanceNow we can use formula 2 to locate the image (S’)

This means the image will be located 1m BEHIND the mirror.This is a VIRTUAL image.

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

m1SS1

2.01

25.01

S1

S1

f1

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:m2.0S

m25.0fm5.0R

focal length

object distanceNow we can use formula 2 to locate the image (S’)

This means the image will be located 1m BEHIND the mirror.This is a VIRTUAL image.

For the magnification, just use formula 3.

5m2.0m1

SSm

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m1SS1

2.01

25.01

S1

S1

f1

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:m2.0S

m25.0fm5.0R

focal length

object distanceNow we can use formula 2 to locate the image (S’)

This means the image will be located 1m BEHIND the mirror.This is a VIRTUAL image.

For the magnification, just use formula 3.

So the image is upright (+) and 5 times as large as the object.

We could also draw the ray diagram…Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

m1SS1

2.01

25.01

S1

S1

f1

5m2.0m1

SSm

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 S S’

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 S<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.

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C

Convex MirrorsThese 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

Rf

3

2

1

object

Image (this is a virtual image behind the mirror, so S’ is negative)

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C F V

Light In SideS > 0 Real Object

Light Out SideS’ > 0 Real ImageC This Side, R > 0

S < 0 Virtual Object

S’ < 0 Virtual ImageC This Side, R < 0

Optic Axis

f1

S1

S1

2Rf S

Syym

C – Center of CurvatureR – Radius of CurvatureF – Focal Point (Same Side as C)V – Vertex

Equations: Paraxial Approximation

Concave Mirror Illustrated

SPHERICAL MIRROR EQUATIONS AND SIGN CONVENTION

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Light In SideS > 0 Real ObjectS’ < 0 Virtual ImageC This Side, R < 0na – Index of Refraction

Light Out SideS < 0 Virtual ObjectS’ > 0 Real ImageC This Side, R > 0nb – Index of Refraction

REFRACTION AT SPHERICAL INTERFACE BETWEEN TWO OPTICAL MATERIALS

Rnn

Sn

Sn abba

SnSn

yym

ba

Illustrated Interface Has C, Center of Curvature, On The Light Out Side, Thus R > 0A Flat Interface Has R = ∞

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Problem 24.23A 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 ZacconeFor Campus Learning Assistance Services at UCSB

Problem 24.23A 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 ZacconeFor 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:

Rnn

S'n

Sn abba

Problem 24.23A 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 ZacconeFor 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:

Rnn

S'n

Sn abba

Here is the given information:cm14R;cm14S;1n;33.1n ba

This radius is negative because the center of the bowl is on the same side as the light source (the fish)

Problem 24.23A 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 ZacconeFor 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:

Rnn

S'n

Sn abba

Here is the given information:cm14R;cm14S;1n;33.1n ba

This radius is negative because the center of the bowl is on the same side as the light source (the fish)cm14Scm14

33.11S1

cm1433.1

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 24.23A 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 ZacconeFor 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:

Rnn

S'n

Sn abba

Here is the given information:cm14R;cm14S;1n;33.1n ba

This radius is negative because the center of the bowl is on the same side as the light source (the fish)cm14Scm14

33.11S1

cm1433.1

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:

SnSnm

ba

Problem 24.23A 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 ZacconeFor 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:

Rnn

S'n

Sn abba

Here is the given information:cm14R;cm14S;1n;33.1n ba

This radius is negative because the center of the bowl is on the same side as the light source (the fish)cm14Scm14

33.11S1

cm1433.1

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:

SnSnm

ba

33.1)cm14(1)cm14(33.1m

The fish appears larger by a factor of 1.33

Problem 24.23A 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 ZacconeFor 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 S’. sunlight

sunlight

Problem 24.23A 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 ZacconeFor 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”.

cm14R;S;33.1n;1n ba 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:

The focal point will be where the sun’s rays converge, so we need to find the image distance S’.

sunlight

Focal Point

Problem 24.23A 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 ZacconeFor 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”.

cm14R;S;33.1n;1n ba This radius is positive because the center of the bowl is on the opposite side as the light source (the sun)cm56Scm14

133.1S33.11

This image is beyond the other side of the bowl (28cm away), so the fish will be safe.

Our given information becomes:

The focal point will be where the sun’s rays converge, so we need to find the image distance S’.

Surface 1 Surface 2

Light In SideS > 0 Real ObjectS’ < 0 Virtual ImageC1 This Side, R1 < 0C2 This Side, R2 < 0

Light Out SideS < 0 Virtual ObjectS’ > 0 Real ImageC1 This Side, R1 > 0C2 This Side, R2 > 0

n – Index of Refraction

C1 – Center of Curvature, Surface 1C2 – Center of Curvature, Surface 2

Illustrated Lens is Double Convex ConvergingWith C1 on the Light Out Side and C2 on the Light In Side

Equations:

SS

yym

f1

S1

S1

21 R1

R11)(n

f1

THIN LENS EQUATIONSAND SIGN CONVENTION

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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 Point

Sample Problema) 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=15cmRadius=20cm

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Sample Problema) 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=15cmRadius=20cm

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

To find the focal length we use the thin lens equation:

21 R1

R11)(n

f1

Sample Problema) 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.

R2=+15cmR1=+20cm

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To find the focal length we use the thin lens equation:

21 R1

R11)(n

f1

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 R1=+20cm.Similarly, the 15cm-radius circle has its center on the other side, so this is also positive: R2=+15cmYour basic rule of thumb is this: follow the light rays – they end up on the positive side.

Sample Problema) 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.

R2=+15cmR1=+20cm

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To find the focal length we use the thin lens equation:

21 R1

R11)(n

f1

cm100fcm151

cm2011)6.1(f

1

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.

Sample Problema) 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.

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For part b) we can use the formula:

f1

S1

S1

f=-100cm S=+50cm

Sample Problema) 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.

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For part b) we can use the formula:

cm33Scm1001

S1

cm501

31

f=-100cm S=+50cm S’=-33.3cm

f1

S1

S1

Sample Problema) 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 ZacconeFor Campus Learning Assistance Services at UCSB

For part b) we can use the formula:

cm33Scm1001

S1

cm501

31

f=-100cm S=+50cm S’=-33.3cm

The height of the image comes from our magnification formula:

cm8ycm50cm33

cm12y

SS

yym 31

The image is virtual, upright, and 8cm tall.

f1

S1

S1

Sample Problema) 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 ZacconeFor Campus Learning Assistance Services at UCSB

For part b) we can use the formula:

cm33Scm1001

S1

cm501

31

f=-100cm S=+50cm S’=-33.3cm

The height of the image comes from our magnification formula:

cm8ycm50cm33

cm12y

SS

yym 31

The image is virtual, upright, and 8cm tall.

f1

S1

S1

Sample Problema) 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 ZacconeFor Campus Learning Assistance Services at UCSB

For part b) we can use the formula:

cm33Scm1001

S1

cm501

31

f=-100cm S=+50cm S’=-33.3cm

The height of the image comes from our magnification formula:

cm8ycm50cm33

cm12y

SS

yym 31

The image is virtual, upright, and 8cm tall.

f1

S1

S1

Sample Problema) 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 ZacconeFor Campus Learning Assistance Services at UCSB

For part b) we can use the formula:

cm33Scm1001

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cm501

31

f=-100cm S=+50cm S’=-33.3cm

The height of the image comes from our magnification formula:

cm8ycm50cm33

cm12y

SS

yym 31

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)

f1

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