A parallel beam of monochromatic light of wavelength 4500 A is incident of YDSE. Let $I_{0}$ is the intensity of incident light. Now two glass slabs $G_{1}$ and $G_{2}$ are kept infront of both slits $S_{1}$ and $S_{2}$ . The refractive index of $G_{1}$ is 1.9 and for $G_{2}$ it is 1.6. The thickness of each glass slab is $15 μ m$ . It is observed that glass $G_{1}$ transmits $\frac{1}{16} t h$ and $G_{2}$ transmits $\frac{1}{25} t h$ of incident energy. If $I_{0} = 8 W a t t / m^{2},$ then find the intensity $(i n W / m^{2})$ observed at point ‘O’ on the screen

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Detailed Solution

Path difference at ‘O’
$= (μ_{1} - μ_{2}) t = 4.5 \times 10^{- 5} m$
Hence phase difference $= ϕ \frac{2 π}{λ} Δ r = 20 π$
Now, $l_{1} = \frac{l_{0}}{16} a n d l_{2} = \frac{l_{0}}{25}$
So, $I = {(\sqrt{l_{1}} + \sqrt{\frac{1}{2}})}^{2}$

$1.62 W / m^{2} = 1.60 W / m^{2}$