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

Hint: 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.