Answer:
d = 0.25 mm
Explanation:
The position that the Minimal fall in single silt diffraction pattern can be written as
y=λmD/d ..........eqn(1)
(y2 - y1 )= difference of y that exist between first and second minima
D= distance of slit from the screen=1.50m
d= width of the slit
λ= 633 nm = 633×10^-9m
y= position of pth minimal
eqn(1) can be written as
(y2 - y1)= λmD/d .........eqn(2)
If we substitute the given values we have
(y2 - y1 )= [ (2×633×10^-9 × 1.50) /d] -[ (1×633×10^-9 × 1.50) /d]
Simplyfying
(y2 - y1)=(9.495×10^-7)/d
But The distance between the first and second minima in the diffraction pattern = 3.75, which implies that (y2 - y1) = 3.75mm=
✓ we can substitute (y2 - y1)= 3.75mm=3.75 ×10^-3 into expression above
3.75 ×10^-3 =(9.495×10^-7)/d
The we can make "d" subject of the formula
d=(9.495×10^-7)/ (3.75 ×10^-3)
d=0.00025m
d=0.25×10^-3 m
d=0.25 mm
Hence, the width (in mm) of the slit is 0.25mm
