Problem 7

  In a Young's double-slit experiment the separation y between the first-order bright fringe and the central bright fringe on a flat screen is 0.0240 m, when the light has a wavelength of 475 nm. Assume that the angles that locate the fringes on the screen are small enough so that $\sin\theta \approx \tan\theta$.Find the separation y when the light has a wavelength of 611 nm.

SOLUTION: From the trigonometry of the problem,

$$\tan\theta = \frac{y}{L},$$

where y is the separation of the central and first-order fringe, and L is the distance from the slits to the screen. Since the angles are small, we're allowed to say that $\tan\theta \approx \sin\theta$.For 1st-order, m = 1, so we have for bright fringes (equation 27.1)

$$\sin\theta = \frac{\lambda}{d},$$

where $\lambda$ is the wavelength of the light, and d is the slit separation. Hence,

$$y \approx \lambda \sin\theta = \lambda \frac{L}{d}.$$

When the light initially has wavelength $\lambda$ = 475 nm, we get

$$\frac{L}{d} = \frac{y}{\lambda} = 5.05 \times 10^4.$$

Plugging this value of into the above equation, we get for the new wavelength of 611 nm,

$$y = (5.05 \times 10^4)(611 \times 10^{-9}) = 0.0309 m.$$