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marusya05 [52]
3 years ago
15

In a double-slit interference experiment, the wavelength is λ = 432 nm, the slit separation is d = 0.100 mm, and the screen is d

= 42.0 cm away from the slits. what is the linear distance between the 8th order maximum and the 4th order maximum on the screen?
Physics
1 answer:
andrew-mc [135]3 years ago
5 0
In the double-slit interference experiment, the distance of the nth-maximum from the center of the screen is given by
y= \frac{n \lambda D}{d}
where
\lambda is the wavelength
D is the distance between the screen and the slits
d is the distance between the slits

In our problem, 
\lambda=432 nm= 432 \cdot 10^{-9} m
D=42.0 cm=0.42 m
d=0.100 mm=0.1 \cdot 10^{-3} m

By applying the previous formula, we can calculate the distance of the 4th maximum from the center of the screen:
y_4 =  \frac{(4)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=7.25 \cdot 10^{-3} m

Similarly, the distance of the 8th- maximum is
y_8 = \frac{(8)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=14.5 \cdot 10^{-3} m

Therefore, the distance between the two maxima is
\Delta y=y_8- y_4 = 14.5 \cdot 10^{-3} m- 7.25 \cdot 10^{-3} m =7.25 \cdot 10^{-3} m = 7.25 mm
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