When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5.When 900-nm light is incident on the same slit system the number is:
A)2
B)3
C)5
D)9
E)10

Question

Quizplus · Accepted Answer

The number of interference maxima within the central diffraction maximum is given by:$$N = \frac{aW}{\lambda}$$where:* N is the number of interference maxima* a is the width of the slits* W is the distance between the slits* λ is the wavelength of lightWhen the wavelength of light is increased, the number of interference maxima decreases. Therefore, when 900-nm light is incident on the same slit system, the number of interference maxima will be:$$N = \frac{aW}{900 	ext{ nm}} = \frac{1}{2} 	imes \frac{aW}{450 	ext{ nm}} = \frac{5}{2} = 2.5$$Since the number of interference maxima must be an integer, the closest integer to 2.5 is 3. Therefore, the correct answer is **C) 5**.

When 450-Nm Light Is Incident Normally on a Certain Double-Slit