In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

Two identical narrow slits $S_1$ and $S_2$ are illuminated by light of wavelength $\lambda$ from a point source $P$. If,as shown in the diagram,the light is then allowed to fall on a screen,and if $n$ is a positive integer,the condition for destructive interference at $Q$ is that

In Young's double-slit experiment,$\frac{d}{D} = 10^{-4}$. The intensity at point $P$ on the screen is equal to the intensity of one of the sources. If the wavelength of the light used is $\lambda = 6000 \, \mathring{A}$,what is the distance of point $P$ from the central bright fringe in $mm$?

Discuss the pattern of interference fringes obtained on the screen away from the two point sources.

In Young's double-slit experiment with a source of light of wavelength $6320 \ \mathring{A}$,the first maxima will occur when:

In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)