The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$

In a Young's double-slit experiment,the slits are $2 \, mm$ apart and are illuminated by a mixture of two wavelengths $\lambda_1 = 7500 \, \mathring{A}$ and $\lambda_2 = 9000 \, \mathring{A}$. At what distance from the central maximum on a screen $2 \, m$ away will a bright fringe from one interference pattern coincide with a bright fringe from the other?

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

The figure shows a double slit experiment where $P$ and $Q$ are the slits. The path lengths $PX$ and $QX$ are $n\lambda$ and $(n + 2)\lambda$ respectively,where $n$ is a whole number and $\lambda$ is the wavelength. Taking the central fringe as zero,what is formed at $X$?

In Young's double slit experiment, the width of the second slit is double the width of the first slit. Consequently, the amplitude of the light from the two slits is different. If $I_m$ is the maximum intensity, the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is given by: