In Young's double-slit experiment how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe if λ = 2000 Å and d = 7000Å. Given that perpendicular distance of a screen from the mid-point of two slits is 3.5 cm.

For,
Wavelength, λ = 2000 Å
Distance between the slits, d = 7000Å
Perpendicular distance of a screen from the mid point of two slits = 3.5 cm

Maximum intensity on the screen

d \sin θ = n λ

i.e.,

\sin θ = \frac{nλ}{d} = \frac{n (2000)}{(7000)} = \frac{n}{3.5}

Since,

\sin θ \leq 1 ∴ n = 0, 1, 2, 3 only

Thus, a total of only seven maximas can be obtained on both sides of the screen.

334 Views

Wave Optics

Hope you found this question and answer to be good. Find many more questions on Wave Optics with answers for your assignments and practice.

Physics Part II

Browse through more topics from Physics Part II for questions and snapshot.