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(B) The fringe width $\beta$ in a Young's double slit experiment is given by the formula:$\beta = \frac{\lambda D}{d}$where $\lambda$ is the wavelengt

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Using blue light instead of green light

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(B) The fringe width $\beta$ in a Young's double slit experiment is given by the formula:$\beta = \frac{\lambda D}{d}$where $\lambda$ is the wavelength of the light used,$D$ is the distance between the slits and the screen,and $d$ is the separation between the two slits.For the fringes to be more closely spaced,the fringe width $\beta$ must decrease.From the formula,$\beta$ is directly proportional to the wavelength $\lambda$ $(\beta \propto \lambda)$.Since the wavelength of blue light is less than the wavelength of green light $(\lambda_{	ext{blue}} < \lambda_{	ext{green}})$,using blue light instead of green light will decrease the fringe width,making the fringes more closely spaced.

In a Young's double slit experiment,green light is incident on the two slits. The interference pattern is observed on a screen. Which of the following changes would cause the observed fringes to be more closely spaced?

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In Young's double-slit experiment,$\frac{d}{D} = 10^{-4}$. The intensity at point $P$ on the screen is equal to the intensity of one of the sources. If the wavelength of the light used is $\lambda = 6000 \, \mathring{A}$,what is the distance of point $P$ from the central bright fringe in $mm$?

The ratio of the distance of the $n^{\text{th}}$ bright band and the $m^{\text{th}}$ dark band from the central bright band in an interference pattern is

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