This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 37 P37.33 393 Treating the anti-reflectance coating like a camera-lens coating, FG
H 2t = m +
Let m = 0 : t= λ
4n = IJ
2n 3.00 cm
= 0.500 cm .
4 1.50 af This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to
1.50 cm—now creating maximum reflection!
H 2nt = m + IJ
2 so Minimum
P37.35 FG 1 IJ λ
H 2 K 2n
F 1 I a500 nmf =
H 2 K 2a1.30f
t= m+ 96.2 nm . Since the light undergoes a 180° phase change at each surface of the film, the condition for
. The film thickness is
constructive interference is 2nt = mλ , or λ =
t = 1.00 × 10 −5 cm = 1.00 × 10 −7 m = 100 nm . Therefore, the wavelengths intensified in the reflected
light are λ= a fa f 2 1.38 100 nm
where m = 1, 2 , 3 , …
m or λ 1 = 276 nm , λ 2 = 138 nm , . . . . All reflection maxima are in the ultraviolet and beyond. No visible wavelengths are intensified.
P37.36 (a) For maximum transmission, we want destructive interference in the light reflected from the
front and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from the front
surface suffers no phase reversal and light reflected from the back does undergo phase
reversal. This effect by itself would produce destructive interference, so we want the
distance down and back to be one whole wavelength in the film: 2t =
2n = λ
n 656.3 nm
= 238 nm
2 1.378 a f (b) The filter will expand. As t increases in 2nt = λ , so does λ increase . (c) Destructive interference for reflected light happens also for λ in 2nt = 2 λ ,
or P37.37 a f λ = 1.378 238 nm = 328 nm anear ultravioletf . If the path length difference ∆ = λ , the transmitted light will be bright. Since ∆ = 2d = λ ,
d min = λ 580 nm
= 290 nm .
View Full Document
This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
- Fall '11