Question

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance between the first and second minima in the diffraction pattern is 3.75 mm .
Required:
What is the width (in mm) of the slit?

Answer 1

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