Question

Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.2m behind the slits. Eleven bright fringes are seen, spanning a distance of 60mm .

What is the spacing (in mm) between the slits?

Answer 1

Step-by-step explanation:

Maximum occurs when the path difference is an integral multiple of wavelength

Here
`\lambda` - Wavelength,
`d-` slit separation and
`m-` Order of pattern

Rearrange the equation for

`d=(m \lambda)/(\left((y)/(L)\right)) \\</p><p>&=(m \lambda L)/(y)`

Here, order

Due to the fact that there are 11 bright fringes seen, you take
`11-1=10`

since starts from 0,1,2,3

Substitute given values

`\begin{aligned}d &=\frac{(10)\left(633 * 10^(-9) \mathrm{m}\right)(3.2 \mathrm{m})}{60 * 10^(-3) \mathrm{m}} \\&=\left(3.376 * 10^(-4) \mathrm{m}\right)\left(\frac{1 \mathrm{mm}}{10^(-3) \mathrm{m}}\right) \\&=0.3376 \mathrm{mm}\end{aligned}`

Answer 2

0.3376 mm

Step-by-step explanation:

The computation of the spacing in mm between the slits is shown below:

As we know that

`d = (m\lambda L)/(\Delta y)`

where,

`\lambda` = wavelength

L = distance from the scrren

`\Delta y` = spanning distance

As there are 11 bright fingers seen so m would be

= 11 - 1

= 10

Now placing these values to the above formula

So, the spacing is

`= ((10)(633 * 10^(-9))(3.2m))/(60 * 10^(-3))`

= 0.3376 mm

We simply applied the above formula.