Question

A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen that is 40 cm away. What is the distance between the two first-order minima?

Answer 1

Data: λ = 550 nm = 550 × 10⁻⁹ m,

a = 0.4 mm = 4 × 10⁻⁴ m

D = 40 cm = 40 × 10⁻² m

`y_md = m (lambda D)/a`

∴ `y_1 d = 1 (lambda D)/a` and 

`2 y_1 d = (2 lambda D)/a`

= `(2 xx 550 xx 10^-9 xx 40 xx 10^-2)/(4 xx 10^-4)`m

= 2 × 550 × 10⁻⁶

= 1100 × 10⁻⁶

= 1.100 × 10⁻³ m

= 1.10 mm

Answer 2

Given:

λ = 550 nm = 550 × 10⁻⁹ m,

a = 0.4 mm = 0.4 × 10⁻³ m

D = 40 cm = 40 × 10⁻² m

To find: Distance between two first order minima

Formula: Width of central maxima, W_c = `(2λ"D")/"a"`

Calculation:

From formula,

W_c = `(2 xx 550 xx 10^-9 xx 40 xx 10^-2)/(0.4 xx 10^-3)`

= 2 × 550 × 10⁻⁶ 

= 1100 × 10⁻⁶

= 1.100 × 10⁻³ m

= 1.10 mm

Distance between two first-order minima = Width of central maxima = 1.10 mm

Distance between two first-order minima is 1.10 mm.