What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?
1 answer:
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Answer:
The minimum thickness = 83.92 nm
Explanation:
The relation between the wavelength in a particular medium and refractive index
where ;
= wavelength of the light in vacuum
n = refractive index of medium with respect to vacuum
For one phase change :
Replacing 1.43 for n and 480 nm for λ; we have:
t = 83.92 nm
Thus; the minimum thickness = 83.92 nm
Answer:
0.75 m
Explanation:
Let's call the distance between the bulb and the mirror x.
The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:
x / 0.4 = (3 + x) / 2
Solving for x:
2x = 0.4 (3 + x)
2x = 1.2 + 0.4 x
1.6 x = 1.2
x = 0.75
So the bulb should located no more than 0.75 m from the mirror.
