In a Young's double slit experiment,the ratio of the slit widths is $4 : 1$. The ratio of the intensity of maxima to minima,close to the central fringe on the screen,will be

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

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:

Light of wavelength $6000 \, Å$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. Find the angular position of the $10^{th}$ maximum in radians.

In a Young's double-slit experiment,sodium light of wavelength $5890 \ \mathring A$ is used,and the angular width of the fringe is found to be $0.20^\circ$. If the angular width is to be increased by $10\%$,what is the required change in the wavelength?

The ratio of the distance of the $n^{\text{th}}$ bright band and the $m^{\text{th}}$ dark band from the central bright band in an interference pattern is