In YDSE,the angular width of a fringe formed | vedclass

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(C) The angular width of a fringe in $YDSE$ is given by $\beta_{	heta} = \frac{\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit sep

Vedclass · Accepted Answer

$0.174\,\mu m$

Vedclass · Answer

(C) The angular width of a fringe in $YDSE$ is given by $\beta_{	heta} = \frac{\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit separation.Given: $\beta_{	heta} = 1^o = \frac{\pi}{180} 	ext{ radians}$ and $d = 0.01\,mm = 10^{-5}\,m$.Substituting the values: $\lambda = \beta_{	heta} 	imes d = \frac{\pi}{180} 	imes 10^{-5}\,m$.$\lambda = 0.01745 	imes 10^{-5}\,m = 1.745 	imes 10^{-7}\,m$.Converting to micrometers: $\lambda = 0.1745 	imes 10^{-6}\,m = 0.1745\,\mu m \approx 0.174\,\mu m$.

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is:

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