1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.49 cm . What is the width of the slit in micrometers (μm)?
???? =_ μm
2).In a double‑slit interference experiment, the wavelength is ????=632 nm , the slit separation is ????=0.100 mm , and the screen is ????=27.0 cm away from the slits. What is the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen?
Δx=___mm

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:

Given as

y = nDλ/w Eqn 1

where

w = width of slit

D = distance to screen

λ = wavelength of light

n = order number

Making x the subject of the formula gives,

w = nDλ/y

Given

y = 0.0149 m

D = 0.555 m

λ = 588 x 10-9 m

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1, y = nDλ/w

given, D = 27cm = 0.27m

λ = 632 x 10-9 m

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm