24 1 Model Each lens is a thin lens The image of the first lens is the object for the second lens Visualize The figure shows the two lenses and a ray tracing diagram The ray tracing shows that the lens combination will produce a real inverted image behind the second lens Solve of the final image is 4 5 cm b s1 15 cm is the object distance of the first lens Its image which is a virtual image is found from the thin lens equation a From the ray tracing diagram we find that the image is 50 cm from the second lens and the height 1 s 1 1 f 1 1 s 1 1 1 40 cm 15 cm 5 120 cm 1 s 24 cm The magnification of the first lens is m 1 s 1 s 1 24 cm 15 cm 1 6 The image of the first lens is now the object for the second lens The object distance is s2 24 cm 10 cm 34 cm A second application of the thin lens equation yields 1 s 2 1 f 2 1 s 2 1 1 20 cm 34 cm 2 s 680 cm 14 48 6 cm The magnification of the second lens is m 2 1 429 s s 2 2 48 6 cm 34 cm The combined magnification is m m m 2 1 6 1 4 57 cm These calculated values are in agreement with those found in part a 1 429 2 286 The height of the final image is 2 286 2 0 cm 24 2 Model Each lens is a thin lens The image of the first lens is the object for the second lens Visualize a From the ray tracing diagram we find that the image is 20 cm in front of the second lens and the The figure shows the two lenses and a ray tracing diagram The ray tracing shows that the lens combination will produce a virtual inverted image in front of the second lens Solve height of the final image is 2 0 cm b s1 60 cm is the object distance of the first lens Its image which is a real image is found from the thin lens equation 1 s 1 1 f s 1 s 1 1 1 40 cm 60 cm 120 cm 1 1 120 cm s The magnification of the first lens is m 1 s 1 s 1 120 cm 60 cm 2 The image of the first lens is now the object for the second lens The object distance is s2 160 cm 120 cm 40 cm A second application of the thin lens equation yields 1 s 2 1 s 2 1 f 2 1 40 cm 1 40 cm 2 s 20 cm The magnification of the second lens is m 2 0 5 s s 2 2 20 cm 40 cm The overall magnification is m m m 2 1 2 0 5 1 0 The height of the final image is 1 0 2 0 cm 2 0 cm The image is inverted because m has a negative sign These calculated values are in agreement with those found in part a 24 3 Model Each lens is a thin lens The image of the first lens is the object for the second lens Visualize a From the ray tracing diagram we find that the image is 10 cm behind the second lens and the height The figure shows the two lenses and a ray tracing diagram The ray tracing shows that the lens combination will produce a real upright image behind the second lens Solve of the final image is 2 cm 20 cm b 1 lens equation is the object distance of the first lens Its image which is real and inverted is found from the thin s 1 s 1 1 s 1 1 f 1 s 1 f s 1 1 f 1 s 1 10 cm 20 cm 20 cm 10 cm 20 cm The magnification of the first lens is m 1 1 s 1 s 1 20 cm 20 cm The s 2 the first 30 cm 20 cm 10 cm image of lens is now the object for A second application of the thin lens equation yields the second lens The object distance is 1 s 2 1 s 2 1 f 2 s 2 f s 2 2 f 2 s 2 5 cm 10 cm 10 cm 5 cm 10 cm The magnification of the second lens is m 2 1 s s 2 2 10 cm 10 cm The combined magnification is These calculated values are in agreement with those found in part a Assess The thin lens equation agrees with the ray tracing 1 1 1 m m m 2 1 The height of the final image is 1 2 0 cm 2 0 cm 24 4 Model Each lens is a thin lens The image of the first lens is the object for the second lens Visualize a From the ray tracing diagram we find that the image is 30 cm in front of the second lens and the The figure shows the two lenses and a ray tracing diagram The ray tracing shows that the lens combination will produce a virtual inverted image at the first lens Solve height of the final image is 6 cm 20 cm b 1 lens equation is the object distance of the first lens Its image which is real and inverted is found from the thin s 1 s 1 1 s 1 1 f 1 s 1 f s 1 1 f 1 s 1 10 cm 20 cm 20 cm 10 cm 20 cm The magnification of the first lens is m 1 1 s 1 s 1 20 cm 20 cm The s 2 the first 30 cm 20 cm 10 cm image of lens is now the object for A second application of the thin lens equation yields the second lens The object distance is 1 s 2 1 s 2 1 f 2 s 2 f s 2 2 f 2 s 2 15 cm 10 cm 10 cm 15 cm 30 cm The magnification of the second lens is m 2 s s 2 2 30 cm 10 cm 3 The height of the final image is 3 2 0 cm 6 0 cm The combined magnification is 3 The image is inverted because m has a negative sign These calculated values are in agreement with those found in part a Assess The thin lens equation agrees with the ray tracing m m m 2 1 3 1 24 5 Model Each lens is a thin lens The image of the first lens is the object for the second lens Visualize The figure shows the two lenses and a ray tracing diagram The ray tracing shows that the lens combination will produce a virtual inverted image in front of the second lens Solve height of the final image is 0 7cm a From the ray tracing diagram we find that the image is …
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