This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Interference Filter - 1 A “one-half-wavelength” thick interference ﬁlter is speciﬁed to pass 500 nm wave—
length light at normal incidence. By rotating the angle of incidence of this ﬁlter it is de-
sired to pass either only 488.0 nm blue light from an argon ion laser or only 514.5 nm '
green light from an argon laser as shown in the ﬁgure. The “one-half—wavelength” thick layers have a refractive index of 1.400. The surrounding regions have a refractive index
of 1.500. Calculate, showing all work, the thickness of the “one—half—wavelength” thick re—
gion. Specify which argon laser wavelength (488.0nm or 514.5 nm) can be passed by
rotating the ﬁlter. Calculate, showing all work, the required angular rotation 04 in de—
grees as measured in air that is required. Express all of your answers accurately to ﬁve signiﬁcant ﬁgures. Put your answers in the space provided. Thickness of “one-half-wavelength” center region 2 nm
The wavelength that can be passed is 488.0 nm 514.5nm. Circle one. Angle of incidence to pass above wavelength (1 = Interference Filter The 500nm wavelength ﬁlter can be tuned to the shorter wavelength of 488.0 nm. The tuning characteristic is
2 n d cos ,6 = m /\ For one half wavelength ﬁlters at normal incidence
2nd = A and so 2nd 2 500nm
and so with n = 1.4 then
d = 178.57 nm
For tuning to 488.0 nm
488.0 nm cosﬁ 500nm and so ,8 578 Parallel glass plates do not affect the ray direction in air. Thus from Snell’s law 1.03ina = 1.4sin,8 and so a 17.7509°. ...
View Full Document
This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Tech.
- Spring '08