In a Young's double slit experiment,the fringe width will remain the same if ($D =$ distance between screen and plane of slits,$d =$ separation between two slits,and $\lambda =$ wavelength of light used).
- ABoth $\lambda$ and $D$ are doubled.
- BBoth $d$ and $D$ are doubled.
- C$D$ is doubled but $d$ is halved.
- D$\lambda$ is doubled but $d$ is halved.
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In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.
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In a certain double-slit experimental arrangement,interference fringes of width $1.0 \ mm$ each are observed when light of wavelength $5000 \ \mathring{A}$ is used. Keeping the setup unaltered,if the source is replaced by another source of wavelength $6000 \ \mathring{A}$,the fringe width will be $...... \ mm$.
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View SolutionSodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:
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