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# g_a - Solution(The Experimental Question Task 1 1a...

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1 0 Solution (The Experimental Question): Task 1 1a. nominal =5´=0.08 nominal (degree) 0.08 1b. If a is the distance between card and the grating and r is the distance between the hole and the light spot so we have ... ,... , 2 2 2 2 1 1 2 1 x x f x x f x x f 2 2 2 0 0 0 0 2 a 2 2 1 , 2 tan a r a r a r If a r We want 0 to be zero i.e. a r r 2 0 0 4 . 0 007 . 0 2 1 70 , 1 0 rad rad a r mm a mm r 0 0.4 range of visible light (degree) 13 26 r Reflected Beam a Optical Axis Incident Beam

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2 1c. 0 min R (21.6±0.1) k 0 = 0.08 1 min R R=(192±1) k 0 =5´ because = 5´ => R= (21.9±0.1) k =- 5´ => R= (21.9±0.1) k 1d. Table 1d. The measured parameters (degree) R glass (M ) R glas s (M ) R film (M ) R film (M ) 15.0 0 3.77 0.03 183 3 15.5 0 2.58 0.02 132 2 16.0 0 1.88 0.01 87 1 16.5 0 1.19 0.01 51.5 0.5 17.0 0 0.89 0.01 33.4 0.3 17.5 0 0.68 0.01 19.4 0.1 18.0 0 0.486 0.005 10.4 0.1 18.5 0 0.365 0.005 5.40 0.03 19.0 0 0.274 0.003 2.66 0. 02 19.5 0 0.225 0.002 1.42 0.01 20.0 0 0.200 0.002 0.880 0.005 20.5 0 0.227 0.002 0.822 0.005 21.0 0 0.368 0.003 1.123 0.007 21.5 0 0.600 0.005 1.61 0.01 22.0 0 0.775 0.005 1.85 0.01 22.5 0 0.83 0.01 1.87 0.01 23.0 0 0.88 0.01 1.93 0.02 23.5 0 1.01 0.01 2. 14 0.02 24.0 0 1.21 0.01 2.58 0.02 24.5 0 1.54 0.01 3.27 0.02 25.0 0 1.91 0.01 4.13 0.02 16.25 1.38 0.01 66.5 0.5 16.75 1.00 0.01 40.0 0.3 17.25 0.72 0.01 23.4 0.2 17.75 0.535 0.005 12.8 0.1 18.25 0.391 0.003 6.83 0.05 18.75 0.293 0.003 3.46 0.02 19 .25 0.235 0.003 1.76 0.01 19.75 0.195 0.002 0.988 0.005 20.25 0.201 0.002 0.776 0.005 20.75 0.273 0.003 0.89 0.01
3 1e.

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