Physics

Solution:

$S_{′}$ represents the another point source forms due to reflection from the mirror.

$S_{′}P=(x+h)_{2}+y_{2} $ and $SP=(x−h)_{2}+y_{2} $

Let path difference at point P be $Δ$

Using, $Δ+SP=S_{′}P$

Squaring and adding, $Δ_{2}+4hx=−2Δ(x−h)_{2}+y_{2} $

Squaring and adding again, $Δ_{4}+16h_{2}x_{2}+8hxΔ_{2}=4Δ_{2}(x_{2}+h_{2}−2hx+y_{2})$

$⟹4(Δ_{2}−4h_{2})x_{2}−16Δ_{2}hx+y_{2}+4Δ_{2}h_{2}−Δ_{4}=0$

which represents equation of circle as $X_{2}+Y_{2}=R_{2}$

Thus the fringes will be of Circular shape centered at O.

A point source is emitting light of wavelength $6000Ao$ is placed at a very small height $h$ above a flat reflecting surface MN as shown in the figure. The intensity of the reflected light is $36%$ of the incident intensity. Inference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $D$ from it. Find the ratio of maximum to minimum intensities at P.

Solution:

Before reflection, intensity of light is $I_{o}$(say) and after reflection it becomes $0.36I_{o}$

$I_{max}=I_{o}+0.36I_{o}+2I_{o}×0.36I_{o} =2.56I_{o}$

$I_{min}=I_{o}+0.36I_{o}−2I_{o}×0.36I_{o} =0.16I_{o}$

So, $I_{min}I_{max} =16256 =16:1$

Solution:

$x=ωλ×D $

$⇒$ $ω=xλ×D $$=5×10_{−3}5000×10_{−10}×2 $$=2×10_{−4}m$$=0.02cm$