An 3.65mm {\ m mm} -tall object is 23.5cm {\ m mm} from the center of a silvered

Question

An 3.65mm { m mm} -tall object is 23.5cm { m mm} from the center of a silvered spherical glass Christmas tree ornament 5.90cm { m mm} in diameter. What is the position of its image (measured from the center of the ornament)? What is the height of its image? When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.30cm { m mm} -tall object is 59.0cm { m mm} to the left of a converging lens of focal length 40.0cm { m mm}. A second converging lens, this one having a focal length of 60.0cm { m mm}, is located 300cm { m mm} to the right of the first lens along the same optic axis. Find the location and height of the image (call it I 1 I_1 is) formed by the lens with a focal length of 40.0cm I 1 I_1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses. BBb{ m mm}

Explanation / Answer

a) f= 40 cm, u = -59 cm

1/v - 1/u = 1/f ;

v = I1 = 124.2 cm right of lens 1.

m = v/u = 124.2/-59 = -2.1

hi = ho*m = -2.73 cm (-ve means it is real and inverted)

b) u = (-300) - (-124.2) = -175.8

f = 60 cm

v = 91.08 right of lens 2

m = v/u = -0.518

hi = ho*m = 1.414 cm (this will be upright, real ; since the object is already inverted.)

Total magnification = 1.414 / 1.3 = 1.0878 (upright)

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