Question

Light of wavelength 755.0 nm passes through a single slit of width 2.20 µm, and a diffraction pattern is observed on a screen 26.5 cm away. Determine the relative intensity of light I/Imax at 13.0 cm away from the central maximum.

Answer 1

The relative intensity of light is
`3.04*10^(-4)`.

Step-by-step explanation:

Given that,

Wavelength = 755.0 nm

Width
`d= 2.20\ \mum`

Distance from screen= 26.5 cm

Distance away from the central maximum = 13.0 cm

Relative intensity of light
`(I)/(I_(max))=13.0\ cm`

We need to calculate the distance

`tan\theta=(y_(n))/(D)`

`\theta=\tan^(-1)(13)/(26.5)`

`\theta=26.13^(\circ)`

The relative intensity of light is

`(I)/(I_(max))=((\sin\phi)/(\phi))^2`

Where,
`\phi=(\pi d\sin\theta)/(\lambda)`

`\phi=(\pi*2.20*10^(-6)*\sin26.13)/(755*10^(-9))`

`\phi=4.032`

Put the value of
`\phi`

`(I)/(I_(max))=((\sin4.032)/(4.032))^2`

`(I)/(I_(max))=3.04*10^(-4)`

Hence, The relative intensity of light is
`3.04*10^(-4)`.

Answer 2

hence the relative intensity is 3.0279 ×
`10^(-4)`

Step-by-step explanation:

Given data

wavelength 755.0 nm = 755 ×
`10^(-9)` m

width d = 2.20 µm = 2.20×
`10^(-6)` m

screen distance D = 26.5 cm

I/Imax Yn = 13.0 cm

to find out

the relative intensity

solution

we know that

tanθ = Yn / D

here Yn is the distance from central maximum and D is here distance from screen put all these value

tanθ = 13 / 26.5

θ = 26.13 degree

so

relative intensity of light will be here

I/ Imax = (sinФ/Ф)² ................1

here Ф is angle that is

Ф = πdsinθ / wavelength

Ф = π(2.20×
`10^(-6)`)sin26.13 / ( 755 ×
`10^(-9)`)

Ф = 7.689

so we put Ф = 7.689 in equation 1

I/ Imax = (sinФ/Ф)²

I/ Imax = (sin7.689/7.689)²

I/ Imax = 3.0279 ×
`10^(-4)`

hence the relative intensity is 3.0279 ×
`10^(-4)`