In an interference experiment,interference fringes are observed on a screen when the chamber is filled with air. If the entire chamber is evacuated and observed carefully,then .....

  • A
    Interference fringes are not observed.
  • B
    Interference fringes become narrower.
  • C
    The intensity of dark fringes increases.
  • D
    The width of the interference fringes increases.

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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:

Obtain the formula for the path difference at a point on the screen in Young's double-slit experiment in terms of $x$,$d$,and $D$.

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is:

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In $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)

In a Young's Double slit experiment,the first maxima is observed at a fixed point $P$ on the screen. Now,the screen is continuously moved away from the plane of the slits. The ratio of the intensity at point $P$ to the intensity at point $O$ (centre of the screen) is:

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