In Young's double slit experiment,the intensity on the screen at a point where path difference is $\lambda$ is $K.$ What will be the intensity at the point where path difference is $\lambda /4$?

$A$ 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:

The intensity of each source in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$

$A$ mixture of light,consisting of wavelengths $590 \ nm$ and an unknown wavelength,illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further,it is observed that the $3^{rd}$ bright fringe of the known light coincides with the $4^{th}$ bright fringe of the unknown light. From this data,the wavelength of the unknown light is ...... $nm$.