Q144 / 160 · TS EAMCET 2019 (03 May Shift 1) · Physics

In a Young's double slit experiment, $m$ th order and $n$th order of bright fringes are formed at point $P$ on a distant screen, if monochromatic source of wavelength $400 \mathrm{~nm}$ and $600 \mathrm{~nm}$ are used respectively. The minimum value of $m$ and $n$ are respectively,

  • A
    4,6
  • Correct answer
    B
    3,2
  • C
    2,3
  • D
    4,2

Solution

Let $m^{\text {th }}$ bright fringe of wavelength $\lambda_{r \pi}$ and $n^{\text {th }}$ bright fringe of wave length $\lambda_n$ formed at a distance $x_n$ from central maxima at same point $P$, i.e., they coincide, therefore, $x_n=x_n$. $$ \begin{aligned} \Rightarrow \quad \frac{m \lambda_{n m} D}{a} & =\frac{n \lambda_n D}{a} \\ \Rightarrow \quad \frac{m}{n} & =\frac{\lambda_n}{\lambda_n} \end{aligned} $$ Putting the given values, we get $$ \begin{aligned} & =\frac{600 \times 10^{-9}}{400 \times 10^{-9}} \\ \text { [Given, } \lambda_m & =400 \times 10^{-9} \mathrm{~m}, \lambda_n=600 \times 10^{-9} \mathrm{~m} \text { ] } \\ \therefore \quad & \frac{m}{n}=\frac{3}{2} \end{aligned} $$ So, least integral values of $m$ and $n$ satisfying above requirement are $m=3$ and $n=2$.

Hence, the correct answer is option: 'B'.

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