According to the formula of diffraction fringe width △x=Ldλ, it can be seen that if the width d of the slit decreases, the fringe spacing △x increases, that is, the diffraction fringe becomes wider.
(2) (1) To map the relationship between the resistance and voltage of a small bulb, it is necessary to continuously adjust the voltage at both ends of the bulb from 0 to the rated voltage of 3.8v. The best reading position of the voltmeter is at least greater than 13 of the full scale, while 3.8V is less than 13 of 15, so only a voltmeter with a range of 5v can be selected, so it is selected.
The rated current of the small bulb I=0.3A, which is greater than the maximum range of A 1, so only the ammeter A2 with the range of 0.6A can be selected.
Since the voltage at both ends of the small bulb needs to be continuously adjusted from 0 to the rated voltage of 3.8v, the sliding rheostat can only be connected in divided voltage, and the smaller the total resistance of the sliding rheostat, the easier it is to adjust, so the sliding rheostat is selected as R 1.
Because the rated voltage of the small bulb is 3.8v, the power supply can only use E2 with electromotive force of 4V;
(2) When the bulb is not working, the voltage at both ends of the bulb is 0, and it can be seen from the figure that the resistance of the bulb is1.5 Ω; When the voltage across the bulb is 3.0V, the resistance of the bulb is11.5Ω. According to P=U2R, the actual power consumed by the light bulb at this time is p = 3.0211.5 = 911.5 = 0.78w. 。
(3) According to Ohm's law of some circuits, I=UR, so the reciprocal of the slope in Figure 3 is the current I, and it can be seen from Figure 3 that the slope is getting smaller and smaller, so the current increases with the increase of voltage, so the power of the bulb P=UI increases with the increase of voltage U, and I=PU, that is, the greater the voltage, the greater the slope, so A is correct.
So choose one?
So the answer to this question is: (1)0. 15 wide.
(2)①A2 V 1 R 1 E2
② 1.5 1 1.5 0.78
③A