Convexlens

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Light can go from a dense medium to a less dense one at an "impossible" angle: e.g in crown glass, what would happen to a ray whose angle of incidence was θ = 60o? e.g. lying on the bottom of a swimming pool looking up what do you see? A prism can be used to show total internal reflection In crown glass, what would the angle of incidence need to be such that the outgoing ray was exactly at 900? 60o 90o 42o 48o Total Internal reflection can occur repeatedly: this is the idea behind fibre optics. If you want to carry a large amount of signal on one carrier, need a very high frequency (roughly, a voice channel needs 10 kHz, so to carry N voice channels, need 10N kHz. How many voice channels can you carry on a 1 MHz radio wave? How many could you carry on red light? Lenses How does a lens form images?. We can build up a lens from a series of prisms We could add a 2nd. prism, to deviate light more, so that two rays go through the same place There are a variety of lens, but essentially they are converging (usually convex) diverging (usually concave) Convex lenses create a real image (i.e. one that can be cast on a screen) The most important quantity for a lens is the focal length f: i.e. how far from the lens do parallel rays get focussed. Concave lenses cause light to diverge, but the rays can be traced back to an (imaginary) focus. Images are formed as either real or virtual: only a convex lens (positive focal length) can form a real image Rules for "ray-tracing" diagrams: a ray which goes through the centre of the lens is not deflected. a ray that is parallel to the axis must go through the focus. a ray that goes through a focus must go parallel to the axis. We can simulate lens on an optical bench The Thin Lens Formula This is the derivation of the "thin-lens" formula. We can use this to find the relation between the distance to the object, the image and the focal length The magnification \color{red}{ M = \frac{{{\rm{image height}}}}{{{\rm{object height}}}} = \frac{{h_i }}{{h_0 }}} (M can be < 1) We have two sets of similar triangles: \color{red}{ \left| {\frac{{h_i }}{{h_0 }}} \right| = \frac{{d_i }}{{d_0 }} = \frac{{d_i - f}}{f}} so \color{red}{ \frac{{d_i - f}}{f} = \frac{{d_i }}{{d_0 }} \Rightarrow \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} The thin lens equation \color{red}{ \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} In words 1 = 1 + 1 focal length object dist. image dist. The surprising thing about this formula is that it always works provided we remember certain conventions: real objects have do > 0 real images have di> 0 virtual ones have di < 0 objects above the axis have ho > 0 images below the axis have hi < 0 convex lenses have f > 0 concave lenses have f <0 convex mirrors have f < 0 concave mirrors have f >0 mirrors have real images on the same side as the object lenses must be "thin": the approx. that is used is sin(θ)= θ e.g suppose we have a lens with f = 20 cm and an 3 cm high object is placed at a distance of 35 cm: where is the image, and how big is it? We can simulate lens on an optical bench The lens-maker's equation: \color{red}{ \frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{R_1 }} + \frac{1}{{R_2 }}} \right)} applies if the lens is "thin" (means that all angles are small enough so that θ = sin(θ)) R₁,R₂ are the radii of curvature of the lens surface: convex surfaces have positive curvature, e.g. suppose R₁ = 10, R₂ = 20, n = 1.4: what is f? 16.6 cm 30.0 cm 12 cm 0.060cm Focal length of a mirror: f = R/2 because if you place a source at the centre the light must be reflected back there. 1 = 1 + 1 f R R We can simulate a mirror on an optical bench e.g. a spoon What happens if you look at the front of the spoon? What is the focal length? What happens if you put an object inside the focal length? What happens if you look at the front of the spoon? Image is real and inverted What happens if you put an object inside the focal length? Image is virtual and upright. Note we are drawing dotted lines to extend the rays through the foci. The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -20 cm? 10 cm? 5 cm? -5 cm? The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -5 cm. The image of your face will be real and inverted orientation real and same orientation virtual and inverted orientation virtual and same orientation Before we can look at more complicated systems of lenses, we need to understand the effects of light as a wave: this is "physical optics"

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which type ofmirrorhas a flat surface?

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How does a lens form images?. We can build up a lens from a series of prisms We could add a 2nd. prism, to deviate light more, so that two rays go through the same place There are a variety of lens, but essentially they are converging (usually convex) diverging (usually concave) Convex lenses create a real image (i.e. one that can be cast on a screen) The most important quantity for a lens is the focal length f: i.e. how far from the lens do parallel rays get focussed. Concave lenses cause light to diverge, but the rays can be traced back to an (imaginary) focus. Images are formed as either real or virtual: only a convex lens (positive focal length) can form a real image Rules for "ray-tracing" diagrams: a ray which goes through the centre of the lens is not deflected. a ray that is parallel to the axis must go through the focus. a ray that goes through a focus must go parallel to the axis. We can simulate lens on an optical bench The Thin Lens Formula This is the derivation of the "thin-lens" formula. We can use this to find the relation between the distance to the object, the image and the focal length The magnification \color{red}{ M = \frac{{{\rm{image height}}}}{{{\rm{object height}}}} = \frac{{h_i }}{{h_0 }}} (M can be < 1) We have two sets of similar triangles: \color{red}{ \left| {\frac{{h_i }}{{h_0 }}} \right| = \frac{{d_i }}{{d_0 }} = \frac{{d_i - f}}{f}} so \color{red}{ \frac{{d_i - f}}{f} = \frac{{d_i }}{{d_0 }} \Rightarrow \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} The thin lens equation \color{red}{ \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} In words 1 = 1 + 1 focal length object dist. image dist. The surprising thing about this formula is that it always works provided we remember certain conventions: real objects have do > 0 real images have di> 0 virtual ones have di < 0 objects above the axis have ho > 0 images below the axis have hi < 0 convex lenses have f > 0 concave lenses have f <0 convex mirrors have f < 0 concave mirrors have f >0 mirrors have real images on the same side as the object lenses must be "thin": the approx. that is used is sin(θ)= θ e.g suppose we have a lens with f = 20 cm and an 3 cm high object is placed at a distance of 35 cm: where is the image, and how big is it? We can simulate lens on an optical bench The lens-maker's equation: \color{red}{ \frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{R_1 }} + \frac{1}{{R_2 }}} \right)} applies if the lens is "thin" (means that all angles are small enough so that θ = sin(θ)) R₁,R₂ are the radii of curvature of the lens surface: convex surfaces have positive curvature, e.g. suppose R₁ = 10, R₂ = 20, n = 1.4: what is f? 16.6 cm 30.0 cm 12 cm 0.060cm Focal length of a mirror: f = R/2 because if you place a source at the centre the light must be reflected back there. 1 = 1 + 1 f R R We can simulate a mirror on an optical bench e.g. a spoon What happens if you look at the front of the spoon? What is the focal length? What happens if you put an object inside the focal length? What happens if you look at the front of the spoon? Image is real and inverted What happens if you put an object inside the focal length? Image is virtual and upright. Note we are drawing dotted lines to extend the rays through the foci. The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -20 cm? 10 cm? 5 cm? -5 cm? The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -5 cm. The image of your face will be real and inverted orientation real and same orientation virtual and inverted orientation virtual and same orientation Before we can look at more complicated systems of lenses, we need to understand the effects of light as a wave: this is "physical optics"

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How does a lens form images?. We can build up a lens from a series of prisms We could add a 2nd. prism, to deviate light more, so that two rays go through the same place There are a variety of lens, but essentially they are converging (usually convex) diverging (usually concave) Convex lenses create a real image (i.e. one that can be cast on a screen) The most important quantity for a lens is the focal length f: i.e. how far from the lens do parallel rays get focussed. Concave lenses cause light to diverge, but the rays can be traced back to an (imaginary) focus. Images are formed as either real or virtual: only a convex lens (positive focal length) can form a real image Rules for "ray-tracing" diagrams: a ray which goes through the centre of the lens is not deflected. a ray that is parallel to the axis must go through the focus. a ray that goes through a focus must go parallel to the axis. We can simulate lens on an optical bench The Thin Lens Formula This is the derivation of the "thin-lens" formula. We can use this to find the relation between the distance to the object, the image and the focal length The magnification \color{red}{ M = \frac{{{\rm{image height}}}}{{{\rm{object height}}}} = \frac{{h_i }}{{h_0 }}} (M can be < 1) We have two sets of similar triangles: \color{red}{ \left| {\frac{{h_i }}{{h_0 }}} \right| = \frac{{d_i }}{{d_0 }} = \frac{{d_i - f}}{f}} so \color{red}{ \frac{{d_i - f}}{f} = \frac{{d_i }}{{d_0 }} \Rightarrow \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} The thin lens equation \color{red}{ \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} In words 1 = 1 + 1 focal length object dist. image dist. The surprising thing about this formula is that it always works provided we remember certain conventions: real objects have do > 0 real images have di> 0 virtual ones have di < 0 objects above the axis have ho > 0 images below the axis have hi < 0 convex lenses have f > 0 concave lenses have f <0 convex mirrors have f < 0 concave mirrors have f >0 mirrors have real images on the same side as the object lenses must be "thin": the approx. that is used is sin(θ)= θ e.g suppose we have a lens with f = 20 cm and an 3 cm high object is placed at a distance of 35 cm: where is the image, and how big is it? We can simulate lens on an optical bench The lens-maker's equation: \color{red}{ \frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{R_1 }} + \frac{1}{{R_2 }}} \right)} applies if the lens is "thin" (means that all angles are small enough so that θ = sin(θ)) R₁,R₂ are the radii of curvature of the lens surface: convex surfaces have positive curvature, e.g. suppose R₁ = 10, R₂ = 20, n = 1.4: what is f? 16.6 cm 30.0 cm 12 cm 0.060cm Focal length of a mirror: f = R/2 because if you place a source at the centre the light must be reflected back there. 1 = 1 + 1 f R R We can simulate a mirror on an optical bench e.g. a spoon What happens if you look at the front of the spoon? What is the focal length? What happens if you put an object inside the focal length? What happens if you look at the front of the spoon? Image is real and inverted What happens if you put an object inside the focal length? Image is virtual and upright. Note we are drawing dotted lines to extend the rays through the foci. The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -20 cm? 10 cm? 5 cm? -5 cm? The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -5 cm. The image of your face will be real and inverted orientation real and same orientation virtual and inverted orientation virtual and same orientation Before we can look at more complicated systems of lenses, we need to understand the effects of light as a wave: this is "physical optics"

What happens if you look at the front of the spoon? What is the focal length? What happens if you put an object inside the focal length?

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Convexmirror

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Objectives: by the end of this you will be able to explain mirrors understand Snell's law and refraction find what happens to light when it hits a transparent medium Draw ray diagrams Make calculations with simple lenses

Lens mirror and prism definitionphysics

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Lens mirror and prism definitionpdf

-20 cm? 10 cm? 5 cm? -5 cm? The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -5 cm. The image of your face will be real and inverted orientation real and same orientation virtual and inverted orientation virtual and same orientation Before we can look at more complicated systems of lenses, we need to understand the effects of light as a wave: this is "physical optics"

What is the refracted angle? 59.40 52.50 23.30 22.50 We have already seen how a single surface refracts. All optical instruments have at least 2 surfaces. A prism deflects light via two successive refractions sin(θ₁) = n sin(θ₂) etc Total Internal Reflection Light can go from a dense medium to a less dense one at an "impossible" angle: e.g in crown glass, what would happen to a ray whose angle of incidence was θ = 60o? e.g. lying on the bottom of a swimming pool looking up what do you see? A prism can be used to show total internal reflection In crown glass, what would the angle of incidence need to be such that the outgoing ray was exactly at 900? 60o 90o 42o 48o Total Internal reflection can occur repeatedly: this is the idea behind fibre optics. If you want to carry a large amount of signal on one carrier, need a very high frequency (roughly, a voice channel needs 10 kHz, so to carry N voice channels, need 10N kHz. How many voice channels can you carry on a 1 MHz radio wave? How many could you carry on red light? Lenses How does a lens form images?. We can build up a lens from a series of prisms We could add a 2nd. prism, to deviate light more, so that two rays go through the same place There are a variety of lens, but essentially they are converging (usually convex) diverging (usually concave) Convex lenses create a real image (i.e. one that can be cast on a screen) The most important quantity for a lens is the focal length f: i.e. how far from the lens do parallel rays get focussed. Concave lenses cause light to diverge, but the rays can be traced back to an (imaginary) focus. Images are formed as either real or virtual: only a convex lens (positive focal length) can form a real image Rules for "ray-tracing" diagrams: a ray which goes through the centre of the lens is not deflected. a ray that is parallel to the axis must go through the focus. a ray that goes through a focus must go parallel to the axis. We can simulate lens on an optical bench The Thin Lens Formula This is the derivation of the "thin-lens" formula. We can use this to find the relation between the distance to the object, the image and the focal length The magnification \color{red}{ M = \frac{{{\rm{image height}}}}{{{\rm{object height}}}} = \frac{{h_i }}{{h_0 }}} (M can be < 1) We have two sets of similar triangles: \color{red}{ \left| {\frac{{h_i }}{{h_0 }}} \right| = \frac{{d_i }}{{d_0 }} = \frac{{d_i - f}}{f}} so \color{red}{ \frac{{d_i - f}}{f} = \frac{{d_i }}{{d_0 }} \Rightarrow \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} The thin lens equation \color{red}{ \frac{1}{f} = \frac{1}{{d_0 }} + \frac{1}{{d_i }}} In words 1 = 1 + 1 focal length object dist. image dist. The surprising thing about this formula is that it always works provided we remember certain conventions: real objects have do > 0 real images have di> 0 virtual ones have di < 0 objects above the axis have ho > 0 images below the axis have hi < 0 convex lenses have f > 0 concave lenses have f <0 convex mirrors have f < 0 concave mirrors have f >0 mirrors have real images on the same side as the object lenses must be "thin": the approx. that is used is sin(θ)= θ e.g suppose we have a lens with f = 20 cm and an 3 cm high object is placed at a distance of 35 cm: where is the image, and how big is it? We can simulate lens on an optical bench The lens-maker's equation: \color{red}{ \frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{R_1 }} + \frac{1}{{R_2 }}} \right)} applies if the lens is "thin" (means that all angles are small enough so that θ = sin(θ)) R₁,R₂ are the radii of curvature of the lens surface: convex surfaces have positive curvature, e.g. suppose R₁ = 10, R₂ = 20, n = 1.4: what is f? 16.6 cm 30.0 cm 12 cm 0.060cm Focal length of a mirror: f = R/2 because if you place a source at the centre the light must be reflected back there. 1 = 1 + 1 f R R We can simulate a mirror on an optical bench e.g. a spoon What happens if you look at the front of the spoon? What is the focal length? What happens if you put an object inside the focal length? What happens if you look at the front of the spoon? Image is real and inverted What happens if you put an object inside the focal length? Image is virtual and upright. Note we are drawing dotted lines to extend the rays through the foci. The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -20 cm? 10 cm? 5 cm? -5 cm? The back of a spoon acts as a convex mirror. The radius of curvature is 10 cm. The focal length is -5 cm. The image of your face will be real and inverted orientation real and same orientation virtual and inverted orientation virtual and same orientation Before we can look at more complicated systems of lenses, we need to understand the effects of light as a wave: this is "physical optics"

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