If fringe width is $0.4 \,mm$, the distance between the fifth bright and third dark band on the same side is: (in $\,mm$)

$A$ beam of light consisting of two wavelengths $6500 \mathring{A}$ and $5200 \mathring{A}$ is used to obtain interference fringes in Young's double-slit experiment. The distance between slits is $2 \text{ mm}$ and the distance of the screen from the slits is $120 \text{ cm}$. What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide (in $\text{ cm}$)?

In Young's double-slit experiment,the distance between the two slits is $0.2 \, mm$ and the screen is $200 \, cm$ away from the slits. If the wavelength of light used is $5000 \, \mathring{A}$,find the distance of the third bright fringe from the central bright fringe in $cm$.

Consider the following statements in the case of Young's double-slit experiment:
$(1)$ $A$ slit is necessary if we use an ordinary extended source of light.
$(2)$ $A$ slit is not necessary if we use an ordinary but well-collimated beam of light.
$(3)$ $A$ slit is not needed if we use a spatially coherent point source of light.
Which of the above statements is true?

In a double-slit experiment,the angular width of the fringes is $0.20^o$ for sodium light $(\lambda = 5890 \ \mathring{A})$. In order to increase the angular width of the fringes by $10\%$,the necessary change in the wavelength is:

In an interference experiment,the distance between a maximum and the adjacent minimum is .......