Q8E - University Physics with Modern Physics

Question

Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

Step-by-Step Solution

Answer

The separation between the two slits is 0.208 mm.

Step 1: Given Data

Wavelength emitted:  $450 \mathrm{nm}$ 

Distance between slit and screen:  $1.80 \mathrm{m}$ 

Distance between dark fringes: $3.90 \mathrm{mm}$ 

Step 2: (a) Concept of Young’s double slit Experiment.

Double Slit experiment is a simple but substantial demonstration which suggests that both light and matter can exhibit characteristics of waves and particles under the constraints of classical definition.

The expression of location of dark bands here is,

${\mathrm{y}}_{\mathrm{m}+\frac{1}{2}}=\mathrm{R}\frac{\left(\mathrm{m}+{\displaystyle \frac{1}{2}}\right)}{\mathrm{d}}\mathrm{\lambda }$                                                                                                ...(i)

Here, ${\mathit{y}}_{\mathbf{m}}$ is the distance of the fringe, λ is the wavelength, m is the order of the fringe, R is the distance between the screen and the slits and d is the slit width.

Step 3: (b) Determination of the separation between the two slits.

The distance between two adjacent dark bands is given as,

$∆y=\frac{R\lambda }{d}$ 

Solve for the slit width and substitute all the values,

$d=\frac{R\lambda }{∆y}$

 

    $$\begin{gathered} =\frac{\left(1.80 \mathrm{m}\right)\left(4.50\times {10}^{-7}\right)}{3.90\times {10}^{-3}} \\ =2.80\times {10}^{-4 }\mathrm{m} \\ =0.208 \mathrm{mm} \end{gathered}$$

Thus, the distance between slits is 0.208 mm.