In a double-slit experiment, the distance between the slits is 0.2 mm, and the distance to the screen is 150 cm. What wavelength (in nm) is needed to have the intensity at a point 1 mm from the central maximum on the screen be 80% of the maximum intensity?
A) 900
B) 700
C) 500
D) 300
E) 600

Question

Quizplus · Accepted Answer

The intensity at a point on the screen in a double-slit experiment is given by:$$I = I_0 \cos^2(\frac{\pi d}{\lambda} \sin	heta)$$where:* $I_0$ is the maximum intensity* $d$ is the distance between the slits* $\lambda$ is the wavelength of the light* $	heta$ is the angle of the point from the central maximumIn this problem, we are given that $d = 0.2$ mm, $L = 150$ cm, and $I = 0.8I_0$. We can use the small-angle approximation $\sin	heta \approx 	an	heta \approx 	heta$ to simplify the equation to:$$I = I_0 \cos^2(\frac{\pi d}{\lambda} 	heta)$$We are also given that the point is 1 mm from the central maximum, so $	heta = \frac{1}{150} = 0.0067$ radians. Substituting all of these values into the equation, we get:$$0.8I_0 = I_0 \cos^2(\frac{\pi (0.2 	imes 10^{-3})}{\lambda} (0.0067))$$Solving for $\lambda$, we get:$$\lambda = \frac{\pi (0.2 	imes 10^{-3})}{\arccos(0.8^{1/2})} = 900 	ext{ nm}$$Therefore, the correct answer is A.

In a Double-Slit Experiment, the Distance Between the Slits Is