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(D) The wavelength $\lambda$ of light in a vacuum is given by $\lambda = \frac{\lambda_0}{n}$,where $\lambda_0$ is the wavelength in vacuum and $n$ is

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The width of the interference fringes increases.

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(D) The wavelength $\lambda$ of light in a vacuum is given by $\lambda = \frac{\lambda_0}{n}$,where $\lambda_0$ is the wavelength in vacuum and $n$ is the refractive index of the medium. Since the refractive index of air $(n_{air} \approx 1.0003)$ is greater than that of a vacuum $(n_{vac} = 1)$,the wavelength $\lambda$ increases when the chamber is evacuated.Since the fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $D$ is the distance to the screen and $d$ is the slit separation,$\beta$ is directly proportional to $\lambda$ $(\beta \propto \lambda)$.Therefore,as the wavelength increases,the width of the interference fringes increases.

In an interference experiment,interference fringes are observed on a screen when the chamber is filled with air. If the entire chamber is evacuated and observed carefully,then .....

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