If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
Answer:
Explanation: NOTHING IN COMMON
Answer:
Explanation:
y1 = 16.2/2 = 8.1 cm
d = 1.22 lambda D/y1
d = 1.22 x 657 x 10^-9 x 23.7/0.081
d = 2.35 x 10^-4 m = 234.52 um
