Question

Two identical radiators have a separation of d = λ/4, where λ is the wavelength of the waves emitted by either source. The initial phase difference between the sources is π/4. Then the intensity on the screen at a distance point situated at an angle θ = 30⁰ from the radiators is (here I₀ is the intensity at that point due to one radiator)

• A. I₀
• B. 2I₀
• C. 3I₀
• D. 4I₀

Answer 1

Answer: (A)

I₀

Answer 2

Initial phase difference $\varphi_{0} = \frac{\pi}{4}$; Phase difference due to path difference $\varphi' = \frac{2\pi}{\lambda}(\Delta)$

where $\Delta = d\sin\theta$

⇒$\varphi' = \frac{2\pi}{\lambda}(d\sin\theta) = \frac{2\pi}{\lambda} \times \frac{\lambda}{4}(\sin 30^{o}) = \frac{\pi}{4}$

Hence total phase difference $\varphi = \varphi_{0} + \varphi' = \frac{\varphi}{4}$.

By using $I = 4I_{0}\cos^{2}(\varphi/2) = 4I_{0}\cos^{2}\left( \frac{\pi/2}{2} \right) = 2I_{0}$.