Question

What is the maximum number of dark fringes that will be produced on both sides of the central maximum if light (? = 640 nm) is incident on a single slit that is 5.09 × 10-6 m wide?(The 3% margin of error does not apply for this question)

Answer 1

Solution :

Wave length = = 640 nm = 640 x 10^-9 m

The condition for dark fringes is d sin = (n+1/2)

=> n = (d sin ) / - 1/2 ; for maximum order, theta = 90 degres ; sin theta = sin 90 = 1

= (5.09 x 10^-6 m) / ( 640 x 10^-9 m) - 0.5 = 7.95

=> number of dark fringes = 7.95 -0.5 = 7.45 = (approximately) 7