Question

A blue laser beam with a wavelength of 450nm illuminates 2 narrow slits each with a width of 0.5micrometers. The opaque strip separating the strips is 15 micrometers wide. The interference pattern is observed on a screen 85cm away.

1. [2] Relative to the optical axis through the center of the slits, at what angle in radians does the 1st interference minima occur?

2. [2] How far from the axis is this (in cm)?

3. [2] At what angle does the 3rd interference maxima (m=3) occur?

4. [2] How far from the axis is this (in cm)?

5. [4] Sketch the pattern (as a function of angle) from 3 maxima below to 3 maxima above the axis.

Answer 1

Given,

$\lambda = 450nm$

Width of the slit $(w) = 0.5\mu m$

Opaque strip separation $= 15\mu m$

Screen distance from the slit $(D)=85 cm=0.85m$

i) Angular distance $(\theta)=\sin^{-1}(\frac{\lambda}{(m+1)d})$

Here m = 1

$(\theta)=\sin^{-1}(\frac{\lambda}{(1+1)d})$

Now, substituting the values,

$(\theta)=\sin^{-1}(\frac{450\times 10^{-9}}{(2)\times (15+0.5)\times 10^{-6}})$

$=\sin^{-1}(0.0145)$

$\simeq 0.83 ^\circ$

ii) first interference minima $=\frac{D\lambda }{2d}$

Now, substituting the values,

$=\frac{0.5\times 0.85\times 450\times 10^{-9}}{2\times (15+0.5)}$

$=0.0123m$

$\simeq1.23cm$

iii)

For m=3,

$\theta = \sin^{-1}(\frac{m\lambda D}{d})$

Now, substituting the values,

$\theta = \sin^{-1}(\frac{3\times 450\times 10^{-9} m}{15.6\times 10^{-6}})$

$=\sin^{-1}(0.087)$

$=4.99^\circ$

$\simeq 5^\circ$

iv) here m = 3

$y=\frac{m \lambda D}{d}$

Now, substituting the values,

$y=\frac{3\times 450\times 10^{-9} m\times 0.85}{15.6\times 10^{-6}}$

$=0.074m$

$\simeq 7.4 cm$

v) The required pattern is given below-