In Young's double slit interference experiment,the slit widths are in the ratio $1 : 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is

In a Young's double slit experiment,light of $500 \ nm$ is used to produce an interference pattern. When the distance between the slits is $0.05 \ mm$,the angular width (in degrees) of the fringes formed on the distant screen is close to $........^o$.

In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

In a Young's double-slit experiment,the distance between the two slits is twice the wavelength of the light used. Find the total number of bright fringes formed on the screen.

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:

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is: