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White coherent light (400 nm-700 nm) is sent through the slits of a Young's double slit experiment (as shown in the figure). The separation between the slits is 1 mm and the screen is 100 cm away from the slits. There is a hole in the screen at a point 1.5 mm away (along the width of the fringes) from the central line. (a) For which wavelength (s) there will be minima at that point ? (b) which wavelength (s) will have a maximum intensity? |
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Answer» For minima (a) `y=(n+(1)/(2))beta.=(n+(1)/(2))(lambdaD)/(d)` `rArr lambda=(1.5xx10^(-3)xx1xx10^(-3))/(1xx(n+(1)/(2))) =(1.5xx10^(-6))/(n+(1)/(2))` `n=1, lambda=(2)/(3)xx1.5xx10^(-6)=1000 nm` `n=2, lambda=(2)/(5)xx1.5xx10^(-6)= 600 nm` `n=3, lambda=(2)/(7)xx1.5xx10^(-6)=428 nm` Putting integral values of `'n'` `n=1 , lambda=1000 nm` `n=2, lambda=600 nm`. `n=3, lambda=428 nm` So only `lambda=428 nm and lambdad=6000 nm,` will have minima at the hole. Hence they will be ABSENT in the LIGHT coming out. (b) `1.5 mm = n beta`. `1.5 mm = n((lambdaD)/(d))` `rArr lambda=(1.5xx10^(-3)xx1xx10^(-3))/(nxx100xx10^(2))` for `n=1.lambda = 1500 nm`. for `n =2.lambda = 750 nm`. for `n=3. lambda=500 nm`. Hence only `lambda=500 nm` will have maximum intensity. |
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