KVPY2010PhysicsWave Optics
In Young's double slit experiment, the distance between the two slits is 0.1 ~mm , the distance between the slits and the screen is 1 ~m and the wavelength of the light used is 600 ~nm . The intensity at a point on the screen is 75 % of the maximum intensity. What is the smallest distance of this point from the central fringe?
Options
- A1.0 ~mm
- B2.0 ~mm
- C0.5 ~mm
- D1.5 ~mm
Correct answer
A. 1.0 ~mm
Step-by-step solution
d =0.1 ~mm , D =1 ~m , =600 ~nm I _ p =75 % of maximum or I _ p =3 I ₀ Where I₀ is the intensity of a single wave now I _ P =3 I ₀= ( I ₀ )²+ ( I ₀ )²+2 I ₀ I ₀ = 3 , also x = yd D now x = 2 3 = 6 y = D 6 ~d = 600 10⁻⁹ 1 6 0.1 10⁻³ or y =1 ~m