1.

In the front of the upper slit of YDSE apparatus, a thin film of a liquid of refractive index 1.40 is placed. It is a hot day and the liquid starts evaporating from the surface. A beam of light at wavelength 560 nm is incident onto the YDSE apparatus and the intensity I at the centre of the screen is monitored. Figure gives intensity I as a function of time t. The intensitychanges because of evaporation from the two sides of the film. Assume that the film is flat and has parallel sides. Also assume that the film's thickness decreases at a constant rate If the maximum intensity is I_(0) then

Answer»

Intensity at `t=0 sec is I_(0)/sqrt(2)`
Intensity at `t=10sec is (3I_(0))/4`
Intensity at `t=5sec is I_(0)/4`
Intensity at `t=5 sec, I_(0)/(2sqrt(2))`

Solution :`I=I_(0)"cos"^(2) (phi)/2`
`Rightarrow "in" t=10sec,Deltaphi=pi/15xx10=(2pi)/3`
`Rightarrow phi=p+(2pi)/(3)`
`I=I_(0)(cos(pi/2+pi/3))^(2)=I_(0)"sin"^(2)pi/3=3/4I_(0)`
`i n t=5sec,Deltaphi=pi/15xx5=pi/3`
`phi=pi+pi/3`
`I=10(cos(pi/2+pi/6))^(2)=I_(0)"sin"^(2)pi/6=I_(0)/4`


Discussion

No Comment Found

Related InterviewSolutions