Dr Sushil Kumar

Quizzes

Theory of Interference Quiz

Last updated on Sunday, August 16th, 2020

Post Reads : 2,186

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. 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

 
 
 
 

2. 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

 

 

 

 
 
 
 

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

 
 
 
 

4. 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}

 

 
 
 
 

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

 
 
 
 

6. 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}

 
 
 
 

7. 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}

 
 
 
 

8. 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)

 
 
 
 

9. 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

 

 

 
 
 
 

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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