Question

In Young's double-slit experiment d//D = 10^(-4) (d = distance between slits, D = distance of screen from the slits) At point P on the screen, resulting intensity is equal to the intensity due to the individual slit I_(0). Then, the distance of point P from the central maximum is (lambda = 6000 Å)

Answer 1

In Young's double slit experiment d/D = 10¯⁴ where d is the distance between slits and D = distance of screen of the slits. at point P on the screen, resulting intensity is equal to the intensity due to the individual slit $I_0$

To find : The distance of point P from the central maximum is...

solution : as intensity at p = intensity at C = $I_0$

maximum intensity, $I_{max}=(\sqrt{I_0}+\sqrt{I_0})^2=4I_0$

using formula, $I=I_{max}cos^2{\frac{\phi}{2}}$

⇒I₀ = 4I₀ cos²Φ/2

⇒1/4 = (1/2)² = [ cosΦ/2 ]²

⇒cosΦ/2 = cosπ/3

⇒Φ = 2π/3

⇒2π∆x/λ = 2π/3

⇒∆x/λ = 1/3

⇒∆x = λ/3

⇒yd/D = λ/3

⇒y = λD/3d

here λ = 6000 A° = 6 × 10^-7 m, d/D = 10¯⁴

y = (6 × 10^-7)/(3 × 10¯⁴) = 2 × 10¯³ m = 2mm

Therefore the distance of point P from the central maximum is 2mm.