Physics | VITEEE 2018 | Discussion Page

In Young's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is (separation between slits is d)

$\sin^{-1}(\frac{\lambda}{d})$

$\sin^{-1}(\frac{\lambda}{2d})$

$\sin^{-1}(\frac{\lambda}{3d})$

$\sin^{-1}(\frac{\lambda}{4d})$

Answer:

Option C

Explanation:

If a is the amplitude of the wave then

$\frac{I_{max}}{4}$ = a² = a² +a² +2aa cosΦ

or cos Φ = - 1/2 or Φ = 2π/3

Corresponding path difference,

Δx = $\frac{\phi\times\lambda}{2\pi}$ = $\frac{(2\pi/3)\times\lambda}{2\pi}$ = λ/3

So, $d\sin\theta=\frac{\lambda}{3}$

or θ = $\sin^{-1}(\frac{\lambda}{3d})$

Discussion Forum

Answer:

Explanation: