Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

  • A
    $1.4 \times 10^{-4} \text{ cm}$
  • B
    $2.7 \times 10^{-4} \text{ cm}$
  • C
    $4.5 \times 10^{-4} \text{ cm}$
  • D
    $6.2 \times 10^{-4} \text{ cm}$

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Similar Questions

In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

In $YDSE$,$16$ fringes occupy a certain region of space with light of wavelength $6000\,\mathring{A}$. If $24$ fringes occupy the same region,the wavelength of the new light is........$\mathring{A}$.

In an interference pattern,the $(n + 4)^{th}$ order blue bright fringe and the $n^{th}$ order red bright fringe coincide at a point. If the wavelengths of red and blue light are $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$ respectively,then the value of $n$ is . . . . . .

For sustained interference fringes in a double-slit experiment,the essential condition$(s)$ is/are:
$1$. Sources must be coherent.
$2$. The intensities of the two sources must be equal.
Here,the correct option$(s)$ is/are:

Explain Young's double-slit experiment,its experimental arrangement,and how it produces an interference pattern.

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