In Young's double slit experiment, two slits are illuminated with a light of wavelength . The line joining A₁P is perpendicular to A₁A₂ as shown in figure. If the first minimum is detected at P, the value of slits separation 'a' will be (D = distance between source and screen)

Step-by-step solution

From the given figure, the path difference at point P is given by: x = A₂P - A₁P Using Pythagoras theorem in right-angled triangle A₁A₂P , we have A₁P = D and A₂P = D^2 + a^2 . So, x = D^2 + a^2 - D For the first minimum to be detected at P, the path difference must be 2 . D^2 + a^2 - D = 2 D^2 + a^2 = D + 2 Squaring both sides, we get: D^2 + a^2 = D^2 + D + ^2 4 a^2 = D + ^2 4 Since the wavelength of light is very small compared to the distance D ( D ), the term ^2 4 is negligibly small and can be ignored. a^2 D a