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

Answer:

hence the relative intensity is  3.0279 ×$10^{-4}$

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 ×$10^{-4}$

Answer 2

Answer:

The relative intensity of light is $3.04\times10^{-4}$.

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 $\dfrac{I}{I_{max}}=13.0\ cm$

We need to calculate the distance

$tan\theta=\dfrac{y_{n}}{D}$

$\theta=\tan^{-1}\dfrac{13}{26.5}$

$\theta=26.13^{\circ}$

The relative intensity of light is

$\dfrac{I}{I_{max}}=(\dfrac{\sin\phi}{\phi})^2$

Where, $\phi=\dfrac{\pi d\sin\theta}{\lambda}$

$\phi=\dfrac{\pi\times2.20\times10^{-6}\times\sin26.13}{755\times10^{-9}}$

$\phi=4.032$

Put the value of $\phi$

$\dfrac{I}{I_{max}}=(\dfrac{\sin4.032}{4.032})^2$

$\dfrac{I}{I_{max}}=3.04\times10^{-4}$

Hence, The relative intensity of light is $3.04\times10^{-4}$.