Question

When 587.9 nm passes through a single slit 0.73 mm wide, it creates a diffraction pattern. (a) What distance away is the wall if the first minimum is 0.86 mm from the central maximum

Answer 1

From Young's single slit experiment, the distance away from the wall will be 1.068 m

Given that 587.9 nm of wavelength of light passes through a single slit 0.73 mm wide, it creates a diffraction pattern.

From the question, the following parameters are given:

The wavelength of the light λ  =  587.9 nm

The width of the slit a = 0.73 mm

Fringe width X = 0.86 mm

The distance away from the wall D = ?

The fringe width is related to the wavelength  of the light source by the equation:

X = Dλ ÷ a

Substitute all the parameters into the formula

0.83 × $10^{-3}$ = 587.9 × $10^{-9}$ D ÷ 0.73 ×

Cross multiply

587.9 × $10^{-9}$ D = 6.278 × $10^{-7}$

make D the subject of the formula

D = 6.278 × $10^{-7}$ ÷  587.9 × $10^{-9}$

D = 1.068 m

Therefore, the distance away from the wall is 1.068 m

Learn more here: brainly.com/question/23801391

Answer 2

If the first minimum is $0.86 mm$ from the central maximum, the distance away is 1.07 meters.

Given the data in the question;

• Wavelength; $\lambda = 587.9nm = 5.879*10^{-7}m$

• Width of slit; $a = 0.73mm = 0.00073m$

• First minimum; $y = 0.86mm = 0.00086m$

• Since its first, order number; $m = 1$

• Distance;  $L = \ ?$

From Thomas Young's single slit experiment:

$\frac{a*y}{L} = m * \lambda$    

Where a is the width of the slit, y is first minimum, L is the distance, m is the order number and λ is the wavelength.

We substitute our values into the equation

$\frac{0.00073m\ *\ 0.00086m}{L} = 1\ *\ ( 5.879*10^{-7}m)\\\\\frac{0.0000006278m^2}{L} = 5.879*10^{-7}m\\\\L = \frac{0.0000006278m^2}{5.879*10^{-7}m} \\\\L = 1.07m$

Therefore, if the first minimum is $0.86 mm$ from the central maximum, the distance away is 1.07 meters.

Learn more: brainly.com/question/16051099