The effects are produced at a given point in space by two waves described by the equations $ {y_1} = {y_m}\sin \omega t $ and $ {y_2} = {y_m}\sin \left( {\omega t + \phi } \right) $ where $ {y_m} $ is the same for both the waves and $ \phi $ is a phase angle. Tick the correct statement among the following
(A) the maximum intensity that can be achieved at a point is twice the intensity of either wave and occurs if $ \phi = 0 $
(B) the maximum intensity that can be achieved at a point is four times the intensity of either wave and occurs if $ \phi = 0 $
(C) the maximum amplitude that can be achieved at the point its twice the amplitude of either wave and occurs at $ \phi = 0 $
(D) When the intensity is zero the net amplitude is zero and at this point $ \phi = \dfrac{\pi }{4} $

Hint : Use the principle of superposition of waves and check the conditions given in the options. The Principle of superposition states that the resultant displacement of a number of waves at a particular point, in a medium is the vector sum of the individual displacements produced by each of the waves at that point.

Complete Step By Step Answer:
 We know from the principle of superposition of waves, the resultant displacement of a number of waves at a particular point, in a medium, is the vector sum of the individual displacements produced by each of the waves at that point.
Now here we have given two waves, $ {y_1} = {y_m}\sin \omega t $ and $ {y_2} = {y_m}\sin \left( {\omega t + \phi } \right) $ where $ {y_m} $ is the same for both the waves.
Hence, the superposed wave at any point becomes, $ y = {y_1} + {y_2} $
Putting the values that becomes, $ y = {y_m}\sin \omega t + {y_m}\sin \left( {\omega t + \phi } \right) $
Now, let’s check option (A) :
 We know that the intensity of a wave is directly proportional to its amplitude $ I \propto {a^2} $ .
Now, for $ \phi = 0 $ the superposed wave becomes,
 $ y = {y_m}\sin \omega t + {y_m}\sin \omega t = 2{y_m}\sin \omega t $ .
Hence, amplitude of the superposed wave is, $ 2{y_m} $
Therefore intensity of the wave will be proportional to $ I \propto {(2{y_m})^2} \propto 4{y_m}^2 $
Hence, option (A ) is incorrect.
Let’s check option ( B): Since, we can already see that the intensity of the superposed wave at $ \phi = 0 $ is $ I \propto 4{y_m}^2 $ i.e. four times the amplitude of each wave.
Hence, option (B ) is correct.
Let’s check option (C ):
We can see that at amplitude of the wave $ y $ will be maximum when $ {y_m}\sin \omega t + {y_m}\sin \left( {\omega t + \phi } \right) $ will have maximum amplitude. That is only possible if $ \phi = 0 $ . For $ \phi = 0 $ maximum amplitude of the wave is $ 2{y_m} $ .
Hence, the maximum amplitude that can be achieved at the point its twice the amplitude of either wave and occurs at $ \phi = 0 $
Hence, option ( C) is correct.
Now, let’s check option (D): Since, we know intensity of a wave is directly proportional to its amplitude $ I \propto {a^2} $ . Then when intensity will be zero amplitude of the wave will also be zero.
Now, for, $ \phi = \dfrac{\pi }{4} $
 $ y = {y_m}\sin \omega t + {y_m}\sin \left( {\omega t + \dfrac{\pi }{4}} \right) $
Upon simplifying we get,
 $ = {y_m}\sin \omega t + {y_m}(\sin \omega t\cos \dfrac{\pi }{4} + \cos \omega t\sin \dfrac{\pi }{4}) $
 $ = {y_m}\sin \omega t + \dfrac{{{y_m}}}{{\sqrt 2 }}(\sin \omega t + \cos \omega t) $
 $ = \dfrac{{{y_m}}}{{\sqrt 2 }}[(\sqrt 2 + 1)\sin \omega t + \cos \omega t)] $
Hence, we can see that the amplitude of the wave is not zero at $ \phi = \dfrac{\pi }{4} $ .
Hence, option ( D) is incorrect.
Hence, correct options are option (B ), option (D).

Note :
We can directly find the solution of this question using Young’s double slit experiment. The two waves given here are coherent sources upon superposition they will produce interference patterns. We know, condition for constructive interference is given by, phase difference of the waves $ \phi = 2m\pi $ . where, $ m = 0,1,2,3,4..... $ and for destructive interference phase difference of the waves will be $ \phi = (2m - 1)\pi $ where, $ m = 0,1,2,3,4..... $