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This is one of our many quizzes on Wave interference. Click on this to access more such competitive quizzes on Wave Optics.

Wave interference Quiz 2

physics

 Fill up: Two identical metal balls are dropped in a pool of still water in such a way that the distance between the droppage points of the balls is $4\text{ m}$. The type of interference at points on either side of both the balls, greater than $4\text{ m}$ is _____ whereas the type of interference at the midpoint of the distance between the droppage points $(=2\text{ m)}$ is ____. In other words, the phase difference at points on either side of the balls greater than $4\text{ m}$ is an ____ multiple of __ whereas the phase difference at the midpoint of droppage points is an ____ multiple of ____. 

 The concept of different types of interference between two identical waves needs to be considered to fill up the required blanks.

  Two identical metal balls are dropped in a pool of still water in such a way that the distance between the droppage points of the balls is $4\text{ m}$. The type of interference at points on either side of both the balls, greater than $4\text{ m}$ is constructive whereas the type of interference at the midpoint of the distance between the droppage points $(=2\text{ m)}$ is destructive. In other words, the phase difference at points on either side of the balls, greater than $4\text{ m}$ is an even multiple of $\pi $ whereas the phase difference at the midpoint of droppage points is an odd multiple of $\pi $.

 If $P=2.5\sin x$ superimpose with $Q=2.5\sin \left( x+\dfrac{\pi }{4} \right)$, the product of resultant wave of $P$ and $Q$ with itself for $x=\dfrac{\pi }{2}$ is given by

 Resultant wave of two similar waves during interference is the sum of these waves. Determine the square of this resultant wave and substitute the given value of $x$ to find the required answer. 

Resultant wave during interference of two similar waves is the sum of individual waves. Therefore, $P+Q=2.5\sin x+2.5\sin \left( x+\dfrac{\pi }{4} \right) \\ \Rightarrow P+Q=2(2.5)\sin \left( \dfrac{2x+\dfrac{\pi }{4}}{2} \right)\cos \left( \dfrac{-\dfrac{\pi }{4}}{2} \right) \\ \Rightarrow P+Q=5\sin \left( \dfrac{x}{1}+\dfrac{\pi }{8} \right)\cos \left( -\dfrac{\pi }{8} \right)$ Now, product of this resultant wave with itself is given by ${{\left( P+Q \right)}^{2}}=25{{\sin }^{2}}\left( \dfrac{x}{1}+\dfrac{\pi }{8} \right){{\cos }^{2}}\left( -\dfrac{\pi }{8} \right)$ Now, for $x=\dfrac{\pi }{2}$; ${{\left( P+Q \right)}^{2}}=25{{\sin }^{2}}\left( \dfrac{\pi }{2}+\dfrac{\pi }{8} \right){{\cos }^{2}}\left( -\dfrac{\pi }{8} \right) \\ \Rightarrow {{\left( P+Q \right)}^{2}}=25{{\sin }^{2}}\left( \dfrac{5\pi }{8} \right){{\cos }^{2}}\left( \dfrac{\pi }{8} \right) \\ \therefore {{\left( P+Q \right)}^{2}}=25(\dfrac{1}{2}(1)=0$

 Two plane waves of equal wavelength of $350\text{ nm}$ interfere in such a way that the distance between the points of intersection of these waves is $2\text{ m}$. If the phase difference between these waves is $\dfrac{\pi }{3}$, what is the angular deviation of both these waves? 

 The mathematical expression for phase difference between two similar plane waves in the process of interference can be utilised to determine the angle between these waves.    

The interference between two similar plane waves is given by $\Delta \phi =\dfrac{2\pi x\sin \theta }{\lambda }$ By substituting the given values from the question in the above expression, $\dfrac{\pi }{3}=\dfrac{2\pi (2)\sin \theta }{350\times {{10}^{-9}}} \\ \Rightarrow \sin \theta =\dfrac{350\times {{10}^{-9}}}{3\times 4} \\ \Rightarrow \sin \theta =29.16\times {{10}^{-9}} \\ \therefore \theta ={{\sin }^{-1}}(29.16\times {{10}^{-9}})$

 Choose the correct answer: The difference between resultant waves during constructive as well as destructive interference between two similar cosine waves, in which the second wave has a particular phase difference when compared to the first wave, is always 

 A general expression for both waves according to the given condition is noted down. The mathematical expressions for constructive interference and destructive interference can be utilised to arrive at the required difference.  

Suppose the first cosine wave is given by $A\cos (kx-\omega t)$ and the second cosine with a particular phase difference with respect to the first wave is given by $A\cos (kx-\omega t+\phi )$ Resultant wave of these two waves during interference is given by $P+Q=A\cos (kx-\omega t)+A\cos (kx-\omega t+\phi ) \\ \Rightarrow P+Q=2A\cos \left( \dfrac{\phi }{2} \right)\cos \left( kx-\omega t+\dfrac{\phi }{2} \right)$ Now, if the interference is constructive, this resultant wave gets modified to ${{\left( P+Q \right)}_{constructive}}=2A\cos (kx-\omega t)$ while during destructive interference, this resultant wave gets modified to ${{\left( P+Q \right)}_{destructive}}=0$ This is because $\phi =....,-2\pi ,0,2\pi ,...$ for constructive interference whereas $\phi =....,-3\pi ,-\pi ,\pi ,3\pi ,...$ for destructive interference. Now, ${{\left( P+Q \right)}_{constructive}}-{{\left( P+Q \right)}_{destructive}}=2A\cos (kx-\omega t)$ Clearly, the required difference is twice the first cosine wave.

 A ray of wavelength $0.6\times {{10}^{-6}}\text{ m}$ falls on two small apertures, dislocated by $'d'$m. If the interferogram of this setup produces a sixth order brilliant light at a distance of $1.4\text{ mm}$ from the centre of the pattern, evaluate the value of $'d'$in meters. Assume that the interferogram is produced at a distance of $\text{1}\text{.4 m}$ from these apertures.

 The mathematical expression for double slit interference of waves can be utilised to arrive at the required distance between the apertures used. 

The mathematical expression for double slit interference is given by $n\lambda =d\sin \theta $ Given that $n=6 \\ \lambda =0.6\times {{10}^{-6}}\text{ m} \\ \text{sin}\theta =\dfrac{y}{D}=\dfrac{1.4\times {{10}^{-3}}\text{ }}{1.4} \\ \Rightarrow \sin \theta ={{10}^{-3}}$ Substituting these values in the above equation, $n\lambda =d\sin \theta \\ \Rightarrow 6\times 0.6\times {{10}^{-6}}=d\times {{10}^{-3}} \\ \Rightarrow d=\dfrac{6\times 0.6\times {{10}^{-6}}}{{{10}^{-3}}} \\ \therefore d=3.6\times {{10}^{-3}}\text{ m}$