NTA Abhyas JEE Main2020PhysicsWave OpticsPractice
In YDSE arrangement as shown in the figure, fringes are seen on screen using monochromatic source S having wavelength 3000 A ∘ (in the air). S 1 and S 2 are two slits separated by d = 1 mm and D = 1 m . Left of slits S 1 a n d S 2 medium of refractive index n 1 = 2 is present and to the right of S 1 and S 2 medium of n 2 = 3 2 , is present. A thin slab of thickness 't' is placed in front of S 1 . The refractive index
Correct answer
2
Step-by-step solution
Path difference, ∆ x = n 1   S S 2 + n 2 S 2 P - n 1 S S 1 + n 2 S 1 P - ∫ 0 t n 3 - n 2 d x   = n 1   S S 2 - S S 1 + n 2   S 2 P - S 1 P - ∫ 0 t n 3 d x - n 2 t In order to get central maxima at the centre of the screen o = 2 × 1 × 10 - 3 2 2 × 1 + 0 - 2 t + 3 t 2 0.5 t = 1   μm t = 2   μm