CBSE Class 12-science Answered
Sir Pl answer( ¡¡) one
Asked by jain.pradeep | 01 Mar, 2020, 11:21: AM
Expert Answer
(i) maximum occurs when angular position θ = (n+1/2) λ/a , where n is an integer other than zero, λ is wavelength and a is slit width.
For example if n = 1 , we get first maximum when θ = (3/2) ( λ/a ) . If slit is divided into three parts, we find that from the first two parts
destructive interference takes place and only third part giving constructive inereference.
Hence only one third of slit is contributing for maximum brightness
In similar manner for higher orders with n= 2, 3 .. etc, we find contribution for constrictive interfernce happens
only for one fifth of slit , one sixth of slit etc .
Hence intensity of diffraction fringes decreases as n in creases.
-----------------------------------
minimum occurs for angular position nλ/a and maximum ocurs for angular position [n+(1/2)] (λ/a)
hence central maximum has angular width 2λ/a
First order secondary maximum occurs between angular position (3/2)(λ/a) and 2(λ/a) ,
Hence angular width on both sides of central spot, 2(1/2)(λ/a) = (λ/a)
Hence angular width of central maximum is twice that of the first order secondary maximum
Answered by Thiyagarajan K | 01 Mar, 2020, 02:26: PM
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