Question

The angular position of second bright band on either side of the central bright band in a diffraction pattern due to single slit of width a and wavelength is​

Answer 1

Explanation:

Angular position of first maxima (n=1)

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians 

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians Position of first darle band 

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians Position of first darle band aDλ=0.2×10−31.4×5460×10−10

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians Position of first darle band aDλ=0.2×10−31.4×5460×10−103.822×10−3

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians Position of first darle band aDλ=0.2×10−31.4×5460×10−103.822×10−3∴between two dark band 

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians Position of first darle band aDλ=0.2×10−31.4×5460×10−103.822×10−3∴between two dark band =2×3.822×10−3m

Angular position of first maxima (n=1)θ=(n+21)aλ=(1+21)0.2×10−35460×10−10=4.095×10−3 radians∴ Angular spread =2×4.095×10−3×radians8.19×10−3 radians Position of first darle band aDλ=0.2×10−31.4×5460×10−103.822×10−3∴between two dark band =2×3.822×10−3m=7.644mm

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