Question

A 0.52-mmmm-diameter hole is illuminated by light of wavelength 520 nmnm .

Part A What is the width (in mmmm) of the central maximum on a screen 1.5 mm behind the slit?

Answer 1

Solution:-

Given –

The diameter of the hole = 0.52 mm = 0.52*10^-3 m = 5.2*10^-4 m

Wavelength of light is 520 nm = 520*10^-12 mm

The angle subs tended by the maxima is given as,

sinϴ = mλ / d

The angle subs tended by the first central maxima is given as,

Sinϴ = 1*λ / d

Therefore, the distance of the first dark fringes from the centreline is given by ,

Y = λD / a

Y = (520*10^-12mm * 1.5mm)/ 0.52 mm

Y = 1.5*10^-9 mm

The width of central maximum is, 1.5*10^-9 mm