- #1
p3t3r1
- 33
- 0
Well, here is the problem..
In summer months, the amount of solar energy entering a house should be minimized. Window glass is made energy efficient by applying a coating to maximize reflected light. Light in the midrange of the visible spectrum (at 568 nm) travels into energy efficient window glass. What thickness of the added coating is needed to maximize reflected light and thus minimize transmitted light?
(air, n = 1.0, coating, n= 1.4 , glass n = 1.5)
Book's solution:
Reflection occurs both at the air-coating interface and at the coating-glass interface. In both cases, the reflected light is 180 deg out of phase with the incident light, since both reflections occur at a fast-to-slow boundary. The two reflected rays would therefore be in phase if there were a zero path difference. To produce constructive intereference the path difference must be wavelength /2. In otherwords, the coating thickness t must be wavelength /4 , where wavelength is the wavelength of the light in the coating.
My solution:
To produce constructive intereference the path difference must be 1 wavelength, not a half wavelength. Therefore, the coating thickness, t, must be wavelength /2. This makes sense since for reflection, we have phase change of ray 1 been reflected and inverted, which is same as change in wavelength /2. We also have ray 2 been reflected and inverted, which is same as change in wavelength /2. Ray 2 also travels more distance, (t x 2, which would be one wavelength if t = wavelength /2) Therefore, ray 2 would differ from ray 1 by a path difference of 1 wavelength and they would interfere constructively for max reflection. And for transmission, we have ray 1 transmitted through without reflection, hence without a phase change caused by the inversion. Then we have ray 2 reflected, undergo a phase change of 180 degrees, or a wavelength/2. It would also travel 1 wave length more since the extra distance is t x 2. Therefore, ray 2 would be half a wavelength from ray 1 and they would interfere destructively to minimize.
The problem is though. if this sample problem is wrong.. so would all the exercises and pratice problem answers.. which would really suck.
In summer months, the amount of solar energy entering a house should be minimized. Window glass is made energy efficient by applying a coating to maximize reflected light. Light in the midrange of the visible spectrum (at 568 nm) travels into energy efficient window glass. What thickness of the added coating is needed to maximize reflected light and thus minimize transmitted light?
(air, n = 1.0, coating, n= 1.4 , glass n = 1.5)
Book's solution:
Reflection occurs both at the air-coating interface and at the coating-glass interface. In both cases, the reflected light is 180 deg out of phase with the incident light, since both reflections occur at a fast-to-slow boundary. The two reflected rays would therefore be in phase if there were a zero path difference. To produce constructive intereference the path difference must be wavelength /2. In otherwords, the coating thickness t must be wavelength /4 , where wavelength is the wavelength of the light in the coating.
My solution:
To produce constructive intereference the path difference must be 1 wavelength, not a half wavelength. Therefore, the coating thickness, t, must be wavelength /2. This makes sense since for reflection, we have phase change of ray 1 been reflected and inverted, which is same as change in wavelength /2. We also have ray 2 been reflected and inverted, which is same as change in wavelength /2. Ray 2 also travels more distance, (t x 2, which would be one wavelength if t = wavelength /2) Therefore, ray 2 would differ from ray 1 by a path difference of 1 wavelength and they would interfere constructively for max reflection. And for transmission, we have ray 1 transmitted through without reflection, hence without a phase change caused by the inversion. Then we have ray 2 reflected, undergo a phase change of 180 degrees, or a wavelength/2. It would also travel 1 wave length more since the extra distance is t x 2. Therefore, ray 2 would be half a wavelength from ray 1 and they would interfere destructively to minimize.
The problem is though. if this sample problem is wrong.. so would all the exercises and pratice problem answers.. which would really suck.