In a double-slit experiment, the angular width of a fringe for sodium light $(\lambda = 5890 \, \mathring{A})$ is $0.20^o$. If the fringe width increases by $10\%$, what is the change in the wavelength?

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

In Young's double slit experiment,if the source of light changes from orange to blue,then

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 $\frac{I_1 + I_2}{I_0} = \dots$

Assertion: $A$ white source of light during interference forms only white and black fringes.
Reason: Width of fringe is inversely proportional to the wavelength of the light used.

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is $d$ and $D$ is the distance between the source and the screen. The possible wavelengths used are inversely proportional to