# Lecture 26 slide 10 10 sf example 111 s s f image is

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Unformatted text preview: S′ f S′ M =− S object f image f>0 S>0 S’ > 0 Physics 212 Lecture 26, Slide 10 10 S=f Example 111 += S S′ f image is: at infinity… S′ M =− S object f f>0 S>0 Physics 212 Lecture 26, Slide 11 11 Example 0<S<f 111 += S S′ f image is: virtual upright bigger S′ M =− S image f object S>0 S’ < 0 f>0 Physics 212 Lecture 26, Slide 12 12 Diverging Lens: Consider the case where the shape of the lens is Consider such that light rays parallel to the axis of the lens all diverge but such but appear to come from a common spot a distance f in front of the lens: f Physics 212 Lecture 26, Slide 13 13 Example image is: virtual upright smaller 111 += S S′ f S′ M =− S object f image f<0 S>0 S’<0 Physics 212 Lecture 26, Slide 14 14 Executive Summary - Lenses: Executive S > 2f 2f real inverted smaller 2f > S > f real real inverted bigger f >S>0 virtual upright bigger S >0 virtual upright smaller converging 111 += S S′ f diverging f S′ M =− S f Physics 212 Lecture 26, Slide 15 15 It’s always the same: 111 += S S′ f S′ M =− S You just have to keep the signs straight: The sign conventions S: S’ : f: positive if object is “upstream” of lens positive of positive if image is “downstream” of lens positive positive if converging lens Physics 212 Lecture 26, Slide 16 16 Preflights 2 and 3 80 60 0 80 s′ > 0 40 20 100 Image on screen MUST BE REAL 111 += s s′ f M =− 60 s′ s 40 20 0 Physics 212 Lecture 26, Slide 17 17 Preflight 5 BB “No light from the top of the object can be transmitted through the lens, so it will not be seen. “ “the image is reversed so only the upper half of the image will show” 50 40 30 20 10 “The rays from the bottom half still focus The image 0 is there, but it will be dimmer !! “ Physics 212 Lecture 26, Slide 18 18 image object Cover top half of lens Light from top of object object image Cover top half of lens Light from bottom of object What’s the Point...
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