JEE Main20231 Feb 2023Evening ShiftPhysicsWave OpticsActual
As shown in the figure, in Young's double slit experiment, a thin plate of thickness t = 10 μm and refractive index μ = 1 . 2 is inserted infront of slit S 1 . The experiment is conducted in air ( μ = 1 ) and uses a monochromatic light of wavelength λ = 500 nm . Due to the insertion of the plate, central maxima is shifted by a distance of x β 0 . β 0 is the fringe-width before the in
Correct answer
0
Step-by-step solution
Fringe shift due to slab in Young's double slit experiment is given by ∆ x = D ( μ - 1 ) t d Since, fringe width β 0 = D d λ Then, ∆ x = β 0 μ - 1 t λ Putting the values, we have = 10 × 10 - 6 ( 1 . 2 - 1 ) 5 × 10 - 7   β 0 = 10 - 5 × 0 . 2 5 × 10 - 7 β 0 = 4 β 0 Hence, the value of x = 4 .