Question

Two wavelengths of sodium light $590$ nm and $596$ nm are used, in turn, to study the diffraction taking place at a single slit of aperture $2 \times {10^{ - 4}}\,m$ . The distance between the slit and the screen is $1.5$ m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Answer 1

Hint:This is a problem on Young’s double slit experiment. The maxima are the points where higher intensities are observed. We have to calculate the first maxima and not the central maxima. Use the formula given below:

Formula used:
$a\theta = \left( {n + \dfrac{1}{2}} \right)\lambda $
where $\theta = \dfrac{y}{D}$
Here $a$ is the aperture;
$n$ denotes the number for maxima to be obtained;
$\lambda $ denotes the wavelength of light used.
$y$ is the position of maxima.
$D$ is the distance between the slit and the screen.

Complete step by step answer:
Young’s double slit experiment is used to understand the wave theory of light. In this experiment, two coherent sources of light are placed at a small distance apart. Maxima is the point where the sum of the intensities of two light sources is maximum. Minima is the point where the intensity is minimum.
For maxima other than central maxima:
$a\theta = \left( {n + \dfrac{1}{2}} \right)\lambda $ and $\theta = \dfrac{y}{D}$
$ \Rightarrow a\dfrac{y}{D} = \left( {n + \dfrac{1}{2}} \right)\lambda $
For light of wavelength ${\lambda _1} = 590nm$ let the position of maxima be ${y_1}$
$ \Rightarrow {y_1} = \dfrac{D}{a}\left( {n + \dfrac{1}{2}} \right){\lambda _1}$
$ \Rightarrow {y_1} = \dfrac{3}{2} \times \dfrac{{590 \times {{10}^{ - 9}} \times 1.5}}{{2 \times {{10}^{ - 4}}}}$
${y_1} = 6.64mm$
Similarly, for light with wavelength ${\lambda _2} = 596nm$ let position of maxima be ${y_2}$
$ \Rightarrow {y_2} = \dfrac{D}{a}\left( {n + \dfrac{1}{2}} \right){\lambda _2}$
$ \Rightarrow {y_2} = \dfrac{3}{2} \times \dfrac{{596 \times {{10}^{ - 9}} \times 1.5}}{{2 \times {{10}^{ - 4}}}}$
$ \Rightarrow {y_2} = 6.705mm$
The separation between the positions of the first maxima of the diffraction pattern is
$\Delta y = {y_2} - {y_1}$
$ \Rightarrow \Delta y = 6.705mm - 6.64mm$
$ \therefore \Delta y = 0.065mm$

Hence, the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases is $0.065mm$.

Additional details:
When light passes through narrow slits, it gets diffracted into semi-circular waves. Pure constructive interference occurs when the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light that falls on the screen and gets scattered to form a pattern.

Note: Convert the values in SI units, while solving the equations. There are more than one maxima and their intensities are different. The first maxima and central maxima are different and they have different intensities. Change in wavelength of light changes the maxima and minima points.