In Young's double slit experiment,the seventh maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum,and the same maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $d_1/d_2$ 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 a Young's double-slit experiment with light of wavelength $\lambda$,the separation of slits is $d$ and the distance of the screen is $D$ such that $D >> d >> \lambda$. If the fringe width is $\beta$,the distance from the point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

In a Young's double-slit experiment,white light is used. The distance between the two slits is $b$ and the distance between the slits and the screen is $d$ (where $d >> b$). At the point on the screen directly in front of one of the slits,certain wavelengths are missing. Some of the missing wavelengths are: $(1) \lambda = b^2/d$,$(2) \lambda = 2b^2/d$,$(3) \lambda = b^2/3d$,$(4) \lambda = 2b^2/3d$.

In $YDSE$,$a=2 \, mm$,$D=2 \, m$,and $\lambda=500 \, nm$. Find the distance of the point on the screen from the central maxima where the intensity becomes $50 \%$ of the central maxima (in $\mu m$).