In an experiment, the expected magnification of the microscope is given by the E
ID: 1694643 • Letter: I
Question
In an experiment, the expected magnification of the microscope is given by the Equation :m=(i_1*L)/(O_1*f_2) Eq1
Refer also to Fig. 4.4 for a definition of the components and distances used in Eq. 1. Suppose you obtain the following data. The distance between the object and the objective lens is 12 cm. The distance between the objective lens and the real, inverted image is 37 cm. The focal length of the eyepiece is 6 cm. When viewing the ruled screen, you observe 4.5 magnified, millimeter divisions filling the 70 mm width of the screen. What eye-to-object distance is consistent with this data?
______________ cm
In an experiment, the expected magnification of the microscope is given by the Equation :
m=(i_1*L)/(O_1*f_2) Eq1
Refer also to Fig. 4.4 for a definition of the components and distances used in Eq. 1. Suppose you obtain the following data. The distance between the object and the objective lens is 12 cm. The distance between the objective lens and the real, inverted image is 37 cm. The focal length of the eyepiece is 6 cm. When viewing the ruled screen, you observe 4.5 magnified, millimeter divisions filling the 70 mm width of the screen. What eye-to-object distance is consistent with this data?
______________ cm
Explanation / Answer
First of all, this question is asking the eye-to-object distance. From the figure, we assume that the eye is on the focal length of lens #2. From the question, we know that the distance would be D = 12 + 37 + distance between lens#2 and candle 2 + 6 [cm] Thus, we need to find the distance between lens#2 and candle 2. So we know that the final image is 4.5 times the original. Set the original candle height to be x [cm]. The final largest candle would be 4.5x [cm]. Set it to be Hf = 4.5x Now find the height of candle 2. Set its height to be H2 we know that 12:x = (12+37) : (x+H2) so H2 = 37x/12 Now we know the relationship between candle 2 and final candle. If we want to find the distance between lens#2 and candle 2, we need to first know the distance from lens #2 to final candle. Set Hd to be the distance from lens #2 to final candle. From the figure, we could find that f2 : H2 = (f2+Hd):Hf, where f2 is the focal length of lens #2 Hd = f2*Hf/H2 - f2 = 6*4.5x/(37x/12) - 6 = 6*4.5*12/37 - 6 =102/37 Now set the distance between lens#2 and candle 2 to be L. L:Hd = H2 : Hf L = Hd*H2/Hf = (102/37)*(37x/12)/(4.5x) = (102/37)*(37/12)*(1/4.5) = 17/9 = 1.8889 [cm] Thus, the eye-to-object distance D = 12 + 37 + distance between lens#2 and candle 2 + 6 [cm] = 12 + 37 + L + 6 = 12 + 37 + 1.8889 + 6 = 56.8889 [cm] Check if the answer is right.... I didn't use the information "you observe 4.5 magnified, millimeter divisions filling the 70 mm width of the screen." Since this question has a lot to do with geometry, feel free to ask me for further help.
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