In a Young's double slit experiment,${I_o}$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be

  • A
    ${I_o}\cos \frac{{\pi x}}{\beta }$
  • B
    $4{I_o}{\cos ^2}\frac{{\pi x}}{\beta }$
  • C
    ${I_o}{\cos ^2}\frac{{\pi x}}{\beta }$
  • D
    $\frac{{{I_o}}}{4}{\cos ^2}\frac{{\pi x}}{\beta }$

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In a Young's double slit experiment,a laser light of $560\,nm$ produces an interference pattern with consecutive bright fringes' separation of $7.2\,mm$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1\,mm$. The wavelength of the second light is $......nm$.

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: If Young's double slit experiment is performed in an optically denser medium than air,then the consecutive fringes come closer.
Reason $(R)$: The speed of light reduces in an optically denser medium than air while its frequency does not change.
In the light of the above statements,choose the most appropriate answer from the options given below:

In Young's double-slit experiment,the intensity at a point is $1/4$ of the maximum intensity. The angular position of this point is:

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

How is the interference pattern affected when violet light replaces sodium light?

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