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JIPMER 2009
Exam
JIPMER
Year
2009
Subject
Physics
Chapter
Wave Optics

JIPMER 2009 · Physics · Wave Optics

The maximum number of possible interference maxima for...

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is

Options

  1. Ainfinite
  2. Bfive
  3. Cthree
  4. Dzero

Correct answer

B. five

Step-by-step solution

For possible interference maxima on the screen, the condition is dsin⁡θ=nλ…(i)d \sin \theta=n \lambda\ldots(i) Given : d=d= slit - width =2λ=2 \lambda∴2λsin⁡θ=nλ⇒2sin⁡θ=n\begin{aligned} & \therefore & 2 \lambda \sin \theta & =n \lambda \\ \Rightarrow & & 2 \sin \theta & =n \end{aligned}$ThemaximumvalueofThe maximum value of\sin \thetais1,hence,is 1 , hence,n=2 \times 1=2Thus,Eq.(i)mustbesatisfiedby5integervaluesThus, Eq. (i) must be satisfied by 5 integer valuesi e,-2,-1,0,1,2$. Hence, the maximum number of possible interference maxima is 5 .
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