The ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is

Question

The ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is (A) 1:2:3 (B) 1:4:9 (C) 1:(2/π 2):(3/π 2) (D) 1:(4/9 π 2):(4/25 π 2)

Tardigrade Admin · Accepted Answer

I=I0((sin (α )/α ))2 and α =0,(3 π /2),(5 π /2), ldots ldots ldots , for maxima The ratio of Intensities of successive maxima is 1:((2/3 π ))2:((2/5 π ))2 or 1:(4/9 π 2):(4/25 π 2)

Q. The ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is

$1 : 2 : 3$

$1 : 4 : 9$

$1 : \frac{2}{π ^{2}} : \frac{3}{π ^{2}}$

$1 : \frac{4}{9 π ^{2}} : \frac{4}{25 π ^{2}}$

$I = I_{0} (\frac{s in ( α )}{α})^{2}$ and $α = 0, \frac{3 π}{2}, \frac{5 π}{2}, \dots\dots\dots$ , for maxima
The ratio of Intensities of successive maxima is
$1 : (\frac{2}{3 π})^{2} : (\frac{2}{5 π})^{2}$ or $1 : \frac{4}{9 π ^{2}} : \frac{4}{25 π ^{2}}$

Q. The ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is

1:2:3

1:4:9

1:π22​:π23​

1:9π24​:25π24​

I=I0​(αsin(α)​)2 and α=0,23π​,25π​,……… , for maxima The ratio of Intensities of successive maxima is 1:(3π2​)2:(5π2​)2 or 1:9π24​:25π24​

$1 : 2 : 3$

$1 : 4 : 9$

$1 : \frac{2}{π ^{2}} : \frac{3}{π ^{2}}$

$1 : \frac{4}{9 π ^{2}} : \frac{4}{25 π ^{2}}$

$I = I_{0} (\frac{s in ( α )}{α})^{2}$ and $α = 0, \frac{3 π}{2}, \frac{5 π}{2}, \dots\dots\dots$ , for maxima
The ratio of Intensities of successive maxima is
$1 : (\frac{2}{3 π})^{2} : (\frac{2}{5 π})^{2}$ or $1 : \frac{4}{9 π ^{2}} : \frac{4}{25 π ^{2}}$