Question

(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.

(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture $2\times {10}^{-6}m$. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

Answer 1

(a) The phenomenon of bending of light round the sharp corners of an obstacle and spreading into the regions of the geometrical shadow is called diffraction.
Expression for fringe width


Consider a parallel light beam from a lens is incident on slit AB. As diffraction happens, the pattern is focussed on screen XY with the help of lens $L_2$. We will get a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes known as secondary maxima and minima.

Central Maximum : Each point on the plane wave front AB sends secondary wavelets in all directions. The waves from points equidistant from the centre C kept on the upper and lower half reach point O with zero path difference and so, reinforce each other, making maximum intensity at point O.

Positions and Widths of Secondary Maxima and Minima

Consider a point P on screen at which wavelets moving in a direction making angle $\theta$ with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference similar to BN :

From the right-angled $\Delta$ ANB, we have :

$BN=ABsin\theta$

$BN=asin\theta ...\left(i\right)$

Suppose $BN=\lambda$ and $\theta ={\theta }_1$

Then, the above equation becomes

$\lambda =asin{\theta }_1$,

$sin{\theta }_1=\frac{\lambda }{a}...\left(ii\right)$

Such a point on the screen will be the position of first secondary minimum,

If $BN=2\lambda$ and $\theta ={\theta }_2$, then

$sin{\theta }_2=\frac{2\lambda }{a}...\left(iii\right)$

Such a point on the screen will be the position of second secondary minimum.

In general, for nth minimum at point P,

$sin{\theta }_n=\frac{n\lambda }{a}...\left(iv\right)$

If $y_n$ is the distance of the nth minimum for the centre of the screen, from right-angled $\Delta$ COP, we have :

$tan{\theta }_n=\frac{OP}{CO}$

$tan{\theta }_n=\frac{y_n}{D}...\left(v\right)$

In case ${\theta }_n$ is small, $tan{\theta }_n\approx sin{\theta }_n$

$\therefore$ Equations (iv) and (v) give

$\frac{y_n}{D}=\frac{n\lambda }{a}$

$y_n=\frac{nD\lambda }{a}$

Width of the secondary maximum,

$\beta =y_n-y_{n-1}=\frac{nD\lambda }{a}-\frac{(n-)D\lambda }{a}$

$\beta =\frac{D\lambda }{a}...\left(vi\right)$

$\therefore \beta$ is independent of n, all the secondary maxima are of the same width $\beta$.

IF $BN=\frac{3\lambda }{2}$ and $\theta ={\theta }^{\prime }$, from equation (i), we have :

$\frac{3\lambda }{2}=asin{\theta }_1^{\prime }$

$sin{\theta }_1^{\prime }=\frac{3\lambda }{2}$

In general, for the nth maximum at point P,

$sin{\theta }_n^{\prime }=\frac{(2n+1)\lambda }{2a}...\left(vii\right)$

If $y_n^{\prime }$ is the distance of nth maximum from the centre of the screen, then the angular position of the maximum is given by

$tan{\theta }_n^{\prime }=\frac{y_n}{D}...\left(viii\right)$

In case ${\theta }_n^{\prime }$ is small,

$sin{\theta }_n^{\prime }\approx tan{\theta }_n^{\prime }$

$\therefore y_n^{\prime }=\frac{(2n+1)D\lambda }{2a}...\left(viii\right)$

Width of the secondary minimum,

${\beta }^{\prime }=y_n^{\prime }-y_{n-1}^{\prime }=\frac{(2n+1)D\lambda }{2a}-\frac{(2n-1)D\lambda }{2a}$

${\beta }^{\prime }=\frac{D\lambda }{a}...\left(ix\right)$

Since ${\beta }^{\prime }$ is independent of n, all the secondary minima are of the same width ${\beta }^{\prime }$.

(b) For first maxima of the diffraction pattern we know

$sin\theta =\frac{3\lambda }{2a}$

where a is aperture of slit.

For small value of $\theta ,sin\theta \approx tan\theta =\frac{y}{D}$

Where y is the distance of first minima from central line and D is the distance between the slit and the screen.

So, $y=\frac{3\lambda }{2a}D$

When $\lambda =590$ nm,

$y_1=\frac{3\times 590\times {10}^{-9}}{2\times 2\times {10}^{-6}}\times 1.5$

$y_1=0.66375$ m

When $\lambda =596$ nm,

$y_2=\frac{3\times 596\times {10}^{-9}}{2\times 2\times {10}^{-6}}\times 1.5$

$y_2=0.6705$ m

Separation between the positions of first maxima = $y_2-y_1=0.00675$ m or 6.75 mm