Physics 6C
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 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 UCSB 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 S and the image distance is labeled S’.
Real Object Virtual Image
S S’
Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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.
R R C C
Shiny side
Shiny side
Concave Mirror – R is positive Convex Mirror – R is negativePrepared by Vince Zaccone For 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)
Prepared by Vince Zaccone For 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)
Ray 1 through the center Optical Axis
Focal Point
Prepared by Vince Zaccone For 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)
Ray 1 through the center Optical Axis
Ray 1 reflects Focal directly back Point
Prepared by Vince Zaccone For 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
Ray 2 through Focal the focal point Point
Prepared by Vince Zaccone For 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)
Ray 2 reflects parallel to axis
Optical Axis
Ray 2 through Focal the focal point Point
Prepared by Vince Zaccone For 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)
Focal Optical Axis Point
Ray 3 comes in parallel to axis
Prepared by Vince Zaccone For 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)
Ray 3 reflects through focal point
Focal Optical Axis Point
Ray 3 comes in parallel to axis
Prepared by Vince Zaccone For 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 Optical Axis Image
Object 1 2
3
Prepared by Vince Zaccone For 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?
Prepared by Vince Zaccone For 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.
1) The focal length is half the radius. Remember the sign convention – if the mirror is R concave R is positive. If convex, R is negative. f = 2
Prepared by Vince Zaccone For 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.
1) The focal length is half the radius. Remember the sign convention – if the mirror is R concave R is positive. If convex, R is negative. f = 2 2) This formula relates the object (S) and image (S’) positions to the focal length (f) of the mirror. Here S is always positive for mirrors, and S’ is positive if the image is on the same 1 1 1 side as the object (a REAL image). = + f S S′ To remember this, just follow the light – a real (positive) image will have light rays passing through it.
Prepared by Vince Zaccone For 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.
1) The focal length is half the radius. Remember the sign convention – if the mirror is R concave R is positive. If convex, R is negative. f = 2 2) This formula relates the object (S) and image (S’) positions to the focal length (f) of the mirror. Here S is always positive for mirrors, and S’ is positive if the image is on the same 1 1 1 side as the object (a REAL image). = + f S S′ To remember this, just follow the light – a real (positive) image will have light rays passing through it.
3) The magnification (m) of the image is related Don’t forget the negative sign in this to the relative positions of the object and image. formula. The sign of m tells you if the image is upright (+) or inverted (-) y′ S′ m = = − y S
Prepared by Vince Zaccone For 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? focal length
OK, back to the problem. We have given information: R = + 5.0 m ⇒ f = + .0 25m S = +0.2m object distance
Prepared by Vince Zaccone For 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? focal length
OK, back to the problem. We have given information: R = + 5.0 m ⇒ f = + .0 25m S = +0.2m Now we can use formula 2 to locate the image (S’) object distance 1 1 1 = + f S S′ 1 1 1 = + ⇒ S′ = −1m 0.25 0.2 S′
Prepared by Vince Zaccone For 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? focal length
OK, back to the problem. We have given information: R = + 5.0 m ⇒ f = + .0 25m S = +0.2m Now we can use formula 2 to locate the image (S’) object distance 1 1 1 = + f S S′ 1 1 1 = + ⇒ S′ = −1m 0.25 0.2 S′
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 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? focal length
OK, back to the problem. We have given information: R = + 5.0 m ⇒ f = + .0 25m S = +0.2m Now we can use formula 2 to locate the image (S’) object distance 1 1 1 = + f S S′ 1 1 1 = + ⇒ S′ = −1m 0.25 0.2 S′
This means the image will be located 1m BEHIND the mirror. This is a VIRTUAL image.
For the magnification, just use formula 3.
S′ − 1m m = − = − = +5 S 0.2m
Prepared by Vince Zaccone For 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? focal length
OK, back to the problem. We have given information: R = + 5.0 m ⇒ f = + .0 25m S = +0.2m Now we can use formula 2 to locate the image (S’) object distance 1 1 1 = + f S S′ 1 1 1 = + ⇒ S′ = −1m 0.25 0.2 S′
This means the image will be located 1m BEHIND the mirror. This is a VIRTUAL image.
For the magnification, just use formula 3.
S′ − 1m m = − = − = +5 S 0.2m
So the image is upright (+) and 5 times as large as the object.
We could also draw the ray diagram… Prepared by Vince Zaccone For 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?
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 3 Object Image 1 f S S’ 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. object R f C 3 Image (this is a virtual image behind 1 the mirror, so S’ is negative) 2 Convex Mirror – R is negative Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB SPHERICAL MIRROR EQUATIONS AND SIGN CONVENTION Concave Mirror Illustrated Light In Side S < 0 Virtual Object S > 0 Real Object C F V Optic Axis Light Out Side S’ < 0 Virtual Image S’ > 0 Real Image C This Side, R < 0 C This Side, R > 0 C – Center of Curvature R – Radius of Curvature F – Focal Point (Same Side as C) V – Vertex Equations: Paraxial Approximation R y′ S′ 1 1 1 f = m = = − + = 2 y S S S′ f Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB REFRACTION AT SPHERICAL INTERFACE BETWEEN TWO OPTICAL MATERIALS Light In Side Light Out Side S > 0 Real Object S < 0 Virtual Object S’ < 0 Virtual Image S’ > 0 Real Image C This Side, R < 0 C This Side, R > 0 n – Index of a nb – Index of Refraction Refraction Illustrated Interface Has C, Center of Curvature, On The Light Out Side, Thus R > 0 A Flat Interface Has R = ∞ n n n − n y′ n S′ a + b = b a m = = − a S S′ R y nbS Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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 UCSB Problem 24.23 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? 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: n n n − n a + b = b a S S' R Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? 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: n n n − n a + b = b a S S' R Here is the given information: na = 1.33;nb = ;1 S = +14cm;R = −14cm This radius is negative because the center of the bowl is on the same side as the light source (the fish) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? 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: n n n − n a + b = b a S S' R Here is the given information: na = 1.33;nb = ;1 S = +14cm;R = −14cm This radius is negative because the center of the 1.33 1 1 − 1.33 bowl is on the same side as + = ⇒ S′ = −14cm the light source (the fish) 14cm S′ − 14cm 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? Magnification can be found For part a) consider the fish to from this formula: be the light source, and n S′ calculate the image position m = − a for light rays exiting the bowl. nbS We will be using this formula: n n n − n a + b = b a S S' R Here is the given information: na = 1.33;nb = ;1 S = +14cm;R = −14cm This radius is negative because the center of the 1.33 1 1 − 1.33 bowl is on the same side as + = ⇒ S′ = −14cm the light source (the fish) 14cm S′ − 14cm 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? Magnification can be found For part a) consider the fish to from this formula: be the light source, and n S′ calculate the image position m = − a for light rays exiting the bowl. nbS We will be using this formula: 1.33(−14cm) n n n − n m = − = 1.33 a + b = b a (1 14cm) S S' R The fish appears larger Here is the given information: by a factor of 1.33 na = 1.33;nb = ;1 S = +14cm;R = −14cm This radius is negative because the center of the 1.33 1 1 − 1.33 bowl is on the same side as + = ⇒ S′ = −14cm the light source (the fish) 14cm S′ − 14cm 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? 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 Our given information becomes: na = ;1 nb = 1.33;S = ∞;R = +14cm This radius is positive because the center of the bowl is on the opposite side as the light source (the sun) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Problem 24.23 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? 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 Our given information becomes: na = ;1 nb = 1.33;S = ∞;R = +14cm Focal Point This radius is positive because the 1 1.33 1.33 − 1 center of the bowl is on the opposite + = ⇒ S′ = +56cm side as the light source (the sun) ∞ S′ + 14cm This image is beyond the other side of the bowl (28cm away), so the fish will be safe. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB THIN LENS EQUATIONS AND SIGN CONVENTION Surface 1 Surface 2 Light In Side Light Out Side S > 0 Real Object S < 0 Virtual Object S’ < 0 Virtual Image S’ > 0 Real Image C1 This Side, R 1 < C1 This Side, R 1 > 0 0 C2 This Side, R 2 > 0 C2 This Side, R 2 < 0 n – Index of Refraction C1 – Center of Curvature, Surface 1 C2 – Center of Curvature, Surface 2 Illustrated Lens is Double Convex Converging With C 1 on the Light Out Side and C 2 on the Light In Side Equations: 1 1 1 1 1 1 y′ S′ = (n−1) − + = m = = − ′ y S f R1 R 2 S S f Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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=20cm Radius=15cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. To find the focal length we use Radius=20cm Radius=15cm the thin lens equation: 1 1 1 = (n−1) − f R1 R2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. To find the focal length we use R1=+20cm R2=+15cm the thin lens equation: Light traveling 1 1 1 this direction = (n−1) − f R1 R2 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=+15cm Your basic rule of thumb is this: follow the light rays – they end up on the positive side. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. To find the focal length we use R1=+20cm R2=+15cm the thin lens equation: Light traveling 1 1 1 this direction = (n−1) − f R1 R2 1 1 1 = 1( .6 − 1) − ⇒ f = −100cm f 20cm 15cm 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. f=-100cm S=+50cm For part b) we can use the formula: 1 1 1 + = S S′ f Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. f=-100cm S=+50cm S’=-33.3cm For part b) we can use the formula: 1 1 1 + = S S′ f 1 1 1 + = ⇒ S′ = −33 1 cm 50cm S′ − 100cm 3 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. f=-100cm S=+50cm S’=-33.3cm For part b) we can use the formula: 1 1 1 + = S S′ f 1 1 1 + = ⇒ S′ = −33 1 cm 50cm S′ − 100cm 3 The height of the image comes from our magnification formula: y′ S′ y′ − 33 1 cm m = = − ⇒ = − 3 ⇒ y′ = +8cm y S 12cm 50cm The image is virtual, upright, and 8cm tall. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. f=-100cm S=+50cm S’=-33.3cm For part b) we can use the formula: 1 1 1 + = S S′ f 1 1 1 + = ⇒ S′ = −33 1 cm 50cm S′ − 100cm 3 The height of the image comes from our magnification formula: y′ S′ y′ − 33 1 cm m = = − ⇒ = − 3 ⇒ y′ = +8cm y S 12cm 50cm The image is virtual, upright, and 8cm tall. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. f=-100cm S=+50cm S’=-33.3cm For part b) we can use the formula: 1 1 1 + = S S′ f 1 1 1 + = ⇒ S′ = −33 1 cm 50cm S′ − 100cm 3 The height of the image comes from our magnification formula: y′ S′ y′ − 33 1 cm m = = − ⇒ = − 3 ⇒ y′ = +8cm y S 12cm 50cm The image is virtual, upright, and 8cm tall. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. f=-100cm S=+50cm S’=-33.3cm The red ray in our diagram is For part b) we can use the formula: initially headed for the focal point on the other side of the lens at x=+100cm. 1 1 1 + = The lens deflects it parallel to the S S′ f axis, and we trace it back to find the image (at the intersection 1 1 1 with the other 2 rays) + = ⇒ S′ = −33 1 cm 50cm S′ − 100cm 3 The height of the image comes from our magnification formula: y′ S′ y′ − 33 1 cm m = = − ⇒ = − 3 ⇒ y′ = +8cm y S 12cm 50cm The image is virtual, upright, and 8cm tall. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB