At two points $P$ and $Q$ on the screen in Young's double-slit experiment,waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be

Light consisting of plane waves of wavelengths $\lambda_1 = 8 \times 10^{-5} \ cm$ and $\lambda_2 = 6 \times 10^{-5} \ cm$ generates an interference pattern in Young's double-slit experiment. If $n_1$ denotes the $n_1^{\text{th}}$ dark fringe due to light of wavelength $\lambda_1$ which coincides with the $n_2^{\text{th}}$ bright fringe due to light of wavelength $\lambda_2$,then:

In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)

$A$ beam of light consisting of two wavelengths $6300 \, \mathring{A}$ and $\lambda \, \mathring{A}$ is used to obtain interference fringes in a Young's double-slit experiment. If the $4^{th}$ bright fringe of $6300 \, \mathring{A}$ coincides with the $5^{th}$ dark fringe of $\lambda \, \mathring{A}$,the value of $\lambda$ (in $\mathring{A}$) is:

In Young's double-slit experiment,the intensity at a point is $(1/4)$ of the maximum intensity. The angular position of this point is:

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?