MHT CET202522 Apr 2025Morning ShiftPhysicsWave OpticsActual
In a Young's double slit experiment wavelength of light used is 6000 Å . The first order maxima and tenth order maxima fall at 14.50 mm and 16.75 mm from the particular reference point in the interference pattern respectively. If the wavelength is changed to 5500 Å then the position of zero order and tenth order maxima are respectively [The other arrangements remaining same]
Options
- A14.25 ~mm , 16.55 ~mm
- B12.25 ~mm , 14.55 ~mm
- C10.25 ~mm , 12.55 ~mm
- D16.25 ~mm , 18.55 ~mm
Correct answer
A. 14.25 ~mm , 16.55 ~mm
Step-by-step solution
Position of the central maximum and fringe separation. The position of the n -th order bright fringe in Young's Double Slit Experiment is given by y_n = y₀ + n D d , where y₀ is the position of the central maximum. Define K = D d , so y_n = y₀ + n K . For ₁ = 6000 Å , y₁ = 14.50 mm and y₁₀ = 16.75 mm . Subtracting the equations yields 16.75 - 14.50 = 9 ₁ K so ₁ K = 2.25 9 = 0.25 mm . Substituting into y₁ gives y₀ = 14.50 - 0.25 = 14.25 mm . New wavelength ₂ = 5500 Å : y₀ remains fixed at 14.25 mm . For n=10 , y₁₀'