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In Young's double slit experiment, fringes of certain width are produced on the screen kept at a certain distance from the slits. When the screen is moved away from the slits by 0.1m, fringe width increases by 6xx10^(-5)m. The separation between the slits is 1 mm. calculate the wavelength of the light used. |
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Answer» Solution :`beta=(lamdaD)/(d)`. .(i) `{beta+6xx10^(-5)}=beta.=(lamdaD.)/(d)=(lamda(D+0.1))/(d)`. . (II) EQUATION (ii)-equation (i) gives `6xx10^(-5)=(0.1lamda)/(d)` } As d=1mm `=1xx10^(-3)m`, `lamda=(6xx10^(-5)xxd)/(0.1)=(6xx10^(-5)xx1xx10^(-3))/(10^(-1))}` Arriving `lamda=6xx10^(-7)m=600nm=6000` Å Detailed Answer: `D=0.1m`, `beta=6xx10^(-5)m` `d=1xx10^(-3)m` `beta=(lamdaD)/(d)` `lamda=(betad)/(D)=(6xx10^(-5)xx1xx10^(-3))/(0.1)=(6xx10^(-8))/(0.1)` `=6xx10^(-7)m` |
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