Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the f
lash lasts for a time interval of 1.45×10−3 s with an average light power output of 2.20×105 W .a) If the conversion of electrical energy to light has an efficiency of 90.0 \% (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash?
b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part A. What is the capacitance?
b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part A. What is the capacitance?
1 answer:
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Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>