Question

In a standard YDSE slit width is 1mm and distance of screen from the slit is 1m. If wavelength of light is 632 nm and bright fringe formed at y=1.270mm . Find the difference for the point .

Answer 1

In a standard YDSE slit width, d = 1mm = 10¯³ m and distance of screen from the slit , D = 1m. wavelength of light, λ = 632 nm = 6.32 × 10^-7 m

and bright fringe formed at y = 1.270 mm = 1.27 × 10¯³ m

To find : The path difference for the point.

solution : path difference is given by,

∆x = yd/D

= (1.270 × 10¯³)(10¯³)/(1 )

= 1.270 × 10¯⁶ m

= 1.270 μ m

Therefore the path difference for the point is 1.270 μm.