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:

In a Young's double slit experiment,a student observes $8$ fringes in a certain segment of the screen when a monochromatic light of $600 \ nm$ wavelength is used. If the wavelength of light is changed to $400 \ nm$,then the number of fringes he would observe in the same region of the screen is:

In a Young's double-slit experiment,the fringe width is $0.6 \, mm$ for a wavelength of $4000 \, \mathring{A}$. If the experiment is performed in water,the fringe width becomes ... $mm$. (Refractive index of water $\mu = 1.33$,but assuming the standard physics problem context where $\mu = 1.5$ is often used for glass/water comparison,we will use the provided value $\mu = 1.5$ from the solution).

The maximum intensity 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 in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10d$?

In Young's double-slit experiment,the two slits are $2 \, mm$ apart and are illuminated by a mixture of two wavelengths $\lambda_1 = 12000 \, \mathring{A}$ and $\lambda_2 = 10000 \, \mathring{A}$. At what minimum distance from the common central bright fringe on the screen will a bright fringe from one interference pattern coincide with a bright fringe from the other? The distance between the slits and the screen is $2 \, m$.

In a Young's double slit experiment,the angular width of a fringe is $0.35^{\circ}$ on a screen placed at $2\,m$ away for a particular wavelength of $450\,nm$. The angular width of the fringe,when the whole system is immersed in a medium of refractive index $7/5$,is $\frac{1}{\alpha}$. The value of $\alpha$ is ..............