Question bank/JEE Main/Physics

In the shown figure three slits, separated by
$d = \sqrt{\frac{2\lambda D}{3}}$ and illuminated by monochromatic
parallel beam of light of wavelength $\lambda$. P is a point on the line
perpendicular to the plane of slits through S_3. If I_0 is the
intensity of the wave from each slit, the intensity at p is

A$I_{0}$B${2I}_{0}$C${4I}_{0}$D${3I}_{0}$

Show answer and solution▾

Phase difference between S_2 &
S_3$\ = \frac{2\pi}{\lambda}(\sqrt{D^{2} + d^{2}} - D)$
$= \frac{2\pi}{\lambda}(\frac{D \times d^{2}}{2D^{2}}) = \frac{2\pi}{3}$
Phase difference between S_1 &
S_3$\ = \frac{2\pi}{\lambda} = (\sqrt{D^{2} + 4d^{2}} - D)$
$= \frac{2\pi}{\lambda} \times D(\frac{4d^{2}}{2D}) = \frac{8\pi}{3}$
[IMAGE] Now using phase diagram
$A_{R} = A\sqrt{3}andIR = 3Io$

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx