In Young's double-slit experiment, the separation between the slits is d = 0 . 25 cm and the distance of the screen from the slits is D = 100 cm . When the wavelength of light used in the experiment is λ = 6 0 0 0 Å , the intensity at a distance x = 4 × 10 - 5 m from the central maximum is p q I 0 , where I 0 is the intensity of central maxima and p and q are the smallest integers. What is the value of p + q ?

Step-by-step solution

Path diff. = xd D ⇒  Path diff. = 4 × 10 - 5 × 0.25 × 10 - 2 1 Path diff. = 1 × 1 0 - 7 Phase diff. = path diff. λ × 2 π ⇒  Phase diff. = 1 × 10 - 7 6 × 10 - 7 × 2 π Phase diff. = 2 π 6   ⇒  Phase diff. = π 3  ⇒  ϕ = 6 0 ∘ I R = I 1 + I 2 + 2 I 1 I 2 cos 6 0 ∘  ⇒  I R = I 1 + I 2 + 2 I × 1 2 I R = 3 I  ⇒   I R = 3 I 0 4 ⇒ p = 3 ,   q