In a Young's double-slit experiment using a monochromatic source,what is the shape of the interference fringes formed on a screen?

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 the ideal 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, the fringe width is $0.4 \text{ mm}$. What is the distance between $4^{th}$ dark band and $6^{th}$ bright band on the same side of the interference pattern (in $\text{ mm}$)?

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to: