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:

  • A
    $3.6 \times 10^{-4} \text{ cm}$
  • B
    $3.2 \times 10^{-4} \text{ cm}$
  • C
    $2.4 \times 10^{-4} \text{ cm}$
  • D
    $2.7 \times 10^{-4} \text{ cm}$

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Similar Questions

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

In Young's double-slit experiment,monochromatic light is used to illuminate two slits $A$ and $B$. Interference fringes are obtained on a screen in front of the slits. Now,if a thick glass plate is placed in the path of light from one of the slits,then:

If fringe width is $0.4 \,mm$, the distance between the fifth bright and third dark band on the same side is: (in $\,mm$)

In Young's double slit experiment,carried out with light of wavelength $5000 \ \mathring{A}$,the distance between the slits is $0.3 \ \text{mm}$ and the screen is at $200 \ \text{cm}$ from the slits. The central maximum is at $x=0 \ \text{cm}$. The value of $x$ for the third maxima is ............. $\text{mm}$.

Assertion: In Young's double slit experiment,if the wavelength of incident monochromatic light is doubled,the number of bright fringes on the screen will increase.
Reason: The maximum number of bright fringes on the screen is directly proportional to the wavelength of light used.

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