Question
The maximum numbers of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is
infinite
five
three
zero
Solution
B.
five
The condition of interference maxima is
The magnitude of sin θ lies between 0 and 1
When n = 0, sinθ = 0 ⇒θ = 0°
When n = 1, sin θ = 1/2 ⇒ θ =30°
When n = 2, sin θ = 1⇒ θ =90°
Thus, there is central maximum ( θ = 0°), on other side of it maxima lie at θ = 30° and θ = 90°, so maximum number of possible interference maxima is 5.
