Question

Monochromatic (single-wavelength) light falling on two slits 0.016 mm apart produces the fifth-order bright spot at an 8.8◦angle. What is the wavelength of the light used?

Answer 1

Given Information:

distance = d = 0.016 mm = 0.0016 cm = 0.000016 m

Angle = θ = 8.8°

bright spot = m = 5

Required Information:

Wavelength = λ = ?

Answer:

Wavelength = 4.89x10⁻⁷ m

Step-by-step explanation:

Monochromatic light passes through a double slit. The corresponding diffraction is given by the equation:

dsinθ = mλ

where d is the distance between two slits, m is the order of the diffraction, θ is the angle and λ is the wavelength of the light wave.

λ = dsinθ/m

λ = 0.016*sin(8.8°)/5

λ = 4.89x10⁻⁷ m

Therefore, the wavelength of the monochromatic light falling on two slits is 4.89x10⁻⁷ m

Answer 2

the wavelength of the light used is 4.8 × 10⁻⁷m

Step-by-step explanation:

We have given distance between the two slits

`d=1.6* 10^(-2)mm=1.6* 10^(-5)m`

Angle
`\Theta =8.8^(\circ)`

We know that for nth order fringe

`dsin\Theta =n\lambda`

In question it is given that 5th order so n =5

`So 1.6* 10^(-5)sin\ 8.8 =5\lambda\\\lambda = 4.8* 10^(-7)m`