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

Answer:

0.3376 mm

Explanation:

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

As we know that

$d = \frac{m\lambda L}{\Delta y}$

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

$= \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}$

= 0.3376 mm

We simply applied the above formula.

Answer 2

Answer:

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

$\begin{aligned}d &=\frac{m \lambda}{\sin \theta} \\\text { Here, } \sin \theta &=\frac{y}{L} \quad\left(\begin{array}{l}\text { Here, } L-\text { separation between slit and screen } \\y-\text { Distance between respective fringe from center on screen }\end{array}\right)$

$d=\frac{m \lambda}{\left(\frac{y}{L}\right)} \\&=\frac{m \lambda L}{y}$

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

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