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:

In $Y.D.S.E$, two slits are made $1 \, mm$ apart and the screen is placed $1 \, m$ away. The fringe separation when light of wavelength $500 \, nm$ is used is:

In the double slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is

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 $mm$?

In a $YDSE$,light of two different wavelengths $(\lambda_1, \lambda_2)$ is incident normal to the plane of slits. The $n^{th}$ maxima of $\lambda_1$ coincides with the $m^{th}$ maxima of $\lambda_2$ exactly in front of one of the slits. Given $D = 1.5 \ m$,$d = 3 \ mm$,and $4500 \ \mathring{A} < \lambda_1, \lambda_2 < 7000 \ \mathring{A}$,then $n, m$ and $\lambda_1$ are:

In Young's double slit experiment,the distance between slits and the screen is $1.0\,m$ and monochromatic light of $600\,nm$ is being used. $A$ person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied,the interference pattern disappears for a particular distance $d_0$ between the slits. If the angular resolution of the eye is $\frac{1}{60}^o,$ the value of $d_0$ is close to......$mm$