COMEDK2023Evening ShiftPhysicsWave OpticsActual
In the young's double slit experiment the fringe width of the interference pattern is found to be 3.2 10⁻⁴ ~m , when the light of wave length 6400^ A is used. What will be change in fringe width if the light is replaced with a light of wave length 4800^ A
Options
- A1.6 10⁻⁴ ~m
- B2.4 10⁻⁴ ~m
- C5.6 10⁻⁴ ~m
- D0.8 10⁻⁴ ~m
Correct answer
D. 0.8 10⁻⁴ ~m
Step-by-step solution
The fringe width in Young's double slit experiment is given by the formula = D d , where is the wavelength of light, D is the distance between the screen and the slits, and d is the distance between the two slits. Given the initial state: ₁ = 3.2 10⁻⁴ m ₁ = 6400 For the second state: ₂ = 4800 Since , we have ₂ ₁ = ₂ ₁ . Substituting the values: ₂ = ₁ ₂ ₁ = 3.2 10⁻⁴ 4800 6400 ₂ = 3.2 10⁻⁴ 3 4 = 2.4 10⁻⁴ m The change in fringe width is = ₁ - ₂ = 3.2 10⁻⁴ - 2.4 10⁻⁴ = 0.8 10⁻⁴ m . Answer: 0.8 10⁻⁴ ~m