Question

Ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is
A. $1:2:3$
B. $1:4:9$
C. $1:\dfrac{2}{{{\pi ^2}}}:\dfrac{3}{{{\pi ^2}}}$
D. $1:\dfrac{4}{{9{\pi ^2}}}:\dfrac{4}{{25{\pi ^2}}}$

Answer 1

In this question, we need to find the ratio of intensities in consecutive maxima in a single slit diffraction pattern. The general condition for Constructive Interference in a single slit diffraction pattern is $a sin \theta = \left( {\dfrac{{2n + 1}}{2}} \right)\lambda$ and intensity of single slit diffraction pattern is given by $I = {I_0}{\left[ {\dfrac{{ sin \varphi /2}}{{\varphi /2}}} \right]^2}$ . Using this, we can find the ratio of intensities.
For constructive interference, $a sin \theta = \left( {\dfrac{{2n + 1}}{2}} \right)\lambda$
Where $a$ is the width of the slit
$\lambda$ is the wavelength of the incident light
$I = {I_0}{\left[ {\dfrac{{ sin \varphi /2}}{{\varphi /2}}} \right]^2}$
Where ${I_0}$ is the intensity at $\theta = {0^ \circ }$.

Complete step by step solution:
The general condition for Constructive Interference or ${\text{nth}}$ maxima in a single slit diffraction $a sin \theta = \left( {\dfrac{{2n + 1}}{2}} \right)\lambda$
Intensity at an angle $\theta$ of a single slit diffraction pattern is given by $I = {I_0}{\left[ {\dfrac{{ sin \varphi /2}}{{\varphi /2}}} \right]^2} - - - - - - - \left( 1 \right)$
Here, $\dfrac{\varphi }{2} = \dfrac{{\pi a sin \theta }}{\lambda }$
But we know, $\dfrac{{a sin \theta }}{\lambda } = \left( {\dfrac{{2n + 1}}{2}} \right)$
Substituting this in the above equation we get,
Hence, $\dfrac{\varphi }{2} = \left( {\dfrac{{2n + 1}}{2}} \right)\pi$
Replacing this in equation $\left( 1 \right)$ we get,
$I = {I_0}{\left[ {\dfrac{{ sin \left( {\dfrac{{2n + 1}}{2}} \right)\pi }}{{\left( {\dfrac{{2n + 1}}{2}} \right)\pi }}} \right]^2}$
But ${\left[ { sin \left( {\dfrac{{2n + 1}}{2}} \right)\pi } \right]^2} = 1$
As $\left( {\dfrac{{2n + 1}}{2}} \right)\pi$ is an integral multiple of $\dfrac{\pi }{2}$
$I = \dfrac{{{I_0}}}{{{{\left[ {\left( {\dfrac{{2n + 1}}{2}} \right)\pi } \right]}^2}}}$
For consecutive maxima take the values of $n = 1,2$
Let the intensities of consecutive maximas be ${I_1},{I_2}$
Intensity of central maxima is ${I_0}$
Intensity at the first maxima is ${I_1} = \dfrac{{{I_0}}}{{{{\left[ {\left( {\dfrac{3}{2}} \right)\pi } \right]}^2}}}$
Intensity at the second maxima is ${I_2} = \dfrac{{{I_0}}}{{{{\left[ {\left( {\dfrac{5}{2}} \right)\pi } \right]}^2}}} -$
The ratio of consecutive maximas is given by ${I_0}:{I_1}:{I_2}$
${I_0}:{I_1}:{I_2} = {I_0}:\dfrac{{{I_0}}}{{{{\left[ {\left( {\dfrac{3}{2}} \right)\pi } \right]}^2}}}:\dfrac{{{I_0}}}{{{{\left[ {\left( {\dfrac{5}{2}} \right)\pi } \right]}^2}}}$
On solving we get,
$\Rightarrow {I_0}:{I_1}:{I_2} = 1:\dfrac{4}{{9{\pi ^2}}}:\dfrac{4}{{25{\pi ^2}}}$
Thus, ratio of Intensities of the consecutive maxima is $1:\dfrac{4}{{9{\pi ^2}}}:\dfrac{4}{{25{\pi ^2}}}$
Hence, the correct option is option D.

Note:
That $sin \theta = 0$ , corresponds to the central maxima while $\dfrac{{\pi a sin \theta }}{\lambda } = \pi$ corresponds to the first minima. The general condition for destructive interference is $a sin \theta = n\lambda$ . This equation gives the values of $\theta$ for which the diffraction pattern has zero light intensity—that is, when a dark fringe is formed. However, it tells us nothing about the variation in light intensity along the screen.