Question

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?

Answer 1

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
$\Delta y=y_8- y_4 = 14.5 \cdot 10^{-3} m- 7.25 \cdot 10^{-3} m =7.25 \cdot 10^{-3} m = 7.25 mm$