Dr Sushil Kumar

Quizzes

Theory of Interference Quiz

Last updated on Sunday, August 16th, 2020

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Theory of interference quiz includes basic fundamental quaestions to judge and provide an angle to learn more about the topic.

Interference is a phenomena when two or more than two waves of same frequency with a constant initial phase difference meets or superimpose in a specific region the intensity of these waves distributes in a defined manner. somewhere it becomes maximum and somewhere minimum. In this quiz we will cover some basic concepts of the interference and related numericals.

1. A source is coherent if it satisfy the condition;

 
 
 
 

2. In the Principle of Superposition, mathematical expression where we explain the minima and maxima condition for the interference, we consider mathematical assumptions; one is

a1 + a2 Cosδ = A Cos Φ

what is the second one, select right one and give the answer.

(a) a2 Sinδ = A Sin Φ

(b) a1 + a2 Sinδ = A Sin (Φ +δ)

(c) a2 SinΦ = B Sin δ

(d) a1 + a2 Cosδ = A Cos (Φ +δ)

 
 
 
 

3. The relationship between path difference and phase difference is;

(a) Path difference = 2λ/π x Phase difference

(b) Path difference = 2π/λ x Phase difference

(c) Phase difference = λ/2π x Path difference

(d) Path difference = λ/2π x Phase difference

 
 
 
 

4. The resultant intensity of light at any point is

\fn_phv \large I=a_{1}^{2}+a_{2}^{2}+2a_{1}a_{2}cos\delta

The avaerage intensity is solved as;

\fn_phv \large I_{av}=\frac{\int_{0}^{2\pi }2\pi I d\delta}{\int_{0}^{\pi}d \delta}

 

 

The average intensity result is ;

(a) I1+I2

(b)  I1 – I2

(c) I1 x I2

(d) I1 / I2

 

 

 
 
 
 

5. Angular Fringe Width

angular fringe width

 

The angular fringe width is defined as the angular separation between consecutive bright or dark fringes.

Here tan θ = Fringe width / Distance

because this angle is very small so you can write;

θ=Fringe width / Distance

What will be the result for angular fringe width;

(a) \fn_phv \large \textup{Fringe width}=\frac{2\lambda}{D}

(b) \fn_phv \large \textup{Fringe width}=\frac{\lambda}{2D}

(c) \fn_phv \large \textup{Fringe width}=\frac{\lambda}{2d}

(d) \fn_phv \large \textup{Fringe width}=\frac{2\lambda}{d}

 
 
 
 

6. The coherent sources whose intensity ratio is 81:1 produce interference fringes. Deduce the ratio of maximum intensity to minimum intensity in fringe system.

 
 
 
 

7. In this expression;

I = (a1 – a2)2 + 4 a1a2 Cos2 (δ/2);

a1 and a2 are constants, the intensity depends only on the constant phase difference value that is δ . Then intensity will be minimum if;

(a) Cos2 (δ/2)= +2

(b) Cos δ = +1

(c) Cos δ = -1

(d) Cos δ/2 = -2

 

 

 

 
 
 
 

8. For bright fringe or maximum intensity at the centre (say O), the path difference must be an even multiple of the half-wavelength (λ/2) of light used. In a two slit experiment the distance of the nth bright fringe from the point O will be;

(a) \fn_phv \large y_{n}=\frac{nD \lambda }{2d}

(b) \fn_phv \large y_{n}=\frac{nd \lambda }{2D}

(c) \fn_phv \large y_{n}=\frac{D \lambda }{2d}

(d) \fn_phv \large y_{n}=\frac{2nD }{d \lambda}

 
 
 
 

9. The distance of the nth dark fringe from centre point O is;

(a) \fn_phv \large y_{n}= \left ( n+\frac{1}{2} \right )\frac{D \lambda}{2d}

(b) \fn_phv \large y_{n}= \left ( 2n-\frac{1}{2} \right )\frac{D \lambda}{2d}

(c) \fn_phv \large y_{n}= \left ( n-\frac{1}{2} \right )\frac{d \lambda}{2D}

(d) \fn_phv \large y_{n}= \left ( n-\frac{1}{2} \right )\frac{D \lambda}{2d}

 

 
 
 
 

10. In this expression;

I = (a1 – a2)2 + 4 a1a2 Cos2 (δ/2);

a1 and a2 are constants, the intensity depends only on the constant phase difference value that is δ . Then intensity will be maximum if;

(a) Cos2 (δ/2)= -1

(b) Cos δ = +1

(c) Cos δ = -1

(d) Cos δ/2 = +2

 

 

 

 
 
 
 

Question 1 of 10

READ ALSO: Michelson Interferometer Quiz

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