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Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. calculated the wavelength of another source of light which produces interference fringes separated by 8.1 mm using same arrangement. Also find the minimum value of the order (n) of bright fringes of shorter wavelength which coincides with that of the longer wavelength. |
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Answer» Solution :Here `lamda_(1)=640nm,beta_(1)=7.2mm,and beta_(2)=8.1mm` For same arrangement, `(beta_(2))/(beta_(1))=(lamda_(2))/(lamda_(1))implieslamda_(2)=(beta_(2)lamda_(1))/(beta_(1))=(8.1xx640)/(7.2)=720nm` Let n bright FRINGES of shorter wavelength `(lamda_(1)=640nm)` coincide with (n-1) bright fringes of longer wavelength `(lamda_(2)=720nm)`, then `nlamda_(1)=(n-1)lamda_(2)`. `impliesn=(lamda_(2))/((lamda_(2)-lamda_(1)))=(720)/(720-640)=9`. |
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