Experimental2Solution - Solution to Experimental Question 2...

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Solution to Experimental Question 2 Section 1 i. A typical geometric layout is as shown below. (a) Maximum distance from ruler to screen is advised to increase the spread of the diffraction pattern. (b) Note that the grating (ruler) lines are horizontal, so that diffraction is in the vertical direction. SCREEN LASER 70 mm β RULER FRINGES 1400 mm ii. Vis a vis the diffraction phenomenon, β = ( y 1400 mm ) The angle β is measured using either a protractor (not recommended) or by measuring the value of the fringe separation on the screen, y , for a given order N . If the separation between 20 orders is measured, then N = ± 10 ( N = 0 is central zero order). The values of y should be tabulated for N = 10. If students choose other orders, this is also acceptable. N ± 10 ± 10 ± 10 ± 10 ± 10 ± 10 ± 10 ± 10 ± 10 ± 10 2 y mm 39.0 38.5 39.5 41.0 37.5 38.0 39.0 38.0 37.0 37.5 y mm 19.5 19.25 19.75 20.5 18.75 19.0 19.5 19.0 18.5 18.75 Mean Value = (19 . 25 ± 1 . 25) mm i.e. Mean “spot” distance = 19 . 25 mm for order N = 10. From observation of the ruler itself, the grating period, h = (0 . 50 ± 0 . 02) mm. Thus in the relation = ± h sin β N = 10 h = 0 . 5 mm sin β β = y 1400 mm = 0 . 01375 10 λ = 0 . 006875 mm λ = 0 . 0006875 mm Since β is small, δλ λ δh h + δy y 10% i.e. measured λ = (690 ± 70) nm The accepted value is 680 nm so that the departure from accepted value equals 1 . 5%.
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Section 2 This section tests the student’s ability to make semi-quantitative measurements and the use of judgement in making observations. i. Using the T = 50% transmission disc, students should note that the transmission through the tank is greater than this value. Using a linear approximation, 75% could well be estimated. Using the
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