物理問題一問(波動) [光波]

MM2 89B_HH1412

2013-01-30 8:09 am

波長590nm的黃光法向入射於一塊每亳米刻有400線的衍射光珊。求第三級和第四級亮紋角位置的差別。

A. 13.7°
B. 25.7°
C. 45.1°
D. 70.7°

詳細解釋，唔該。

回答 (1)

天同

2013-01-30 8:32 am

✔ 最佳答案

Use formula: a.sin(p) = m入
where a is the grating line spacing
angle p is the angular position of a bright fringe
入 is the wavelength of light used
m is an interger.

Here, a = 1/400 mm = 2.5 x 10^-3 mm = 2.5 x 10^-6 m
入= 590 x 10^-9 m = 5.9 x 10^-7 m

For the 3rd order fringe, m = 3
Hence, (2.5x10^-6).sin(p) = 3 x 5.9x10^-7
sin(p) = 0.708
p = 45.07 degrees

For the 4th order fringe, m = 4
Hence, (2.5x10^-6).sin(p) = 4 x 5.9x10^-7
sin(p) = 0.944
p = 70.73 degrees

Therefore, separation of angular positions
= (70.73 - 45.07) degrees = 25.66 degrees

The answer is option B

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