An electromagnetic radiation of wavelength ranging between `400nm` and `1150nm` (for which the plate is penetrable) incident perpendicularly on the plate from above is reflected from both the air surfaces and interferes. In this range only two wavelenght give maximum reinforcements, one of them is `lamda=400nm`. (refractive index of air `=1` and the coefficient of linear thermal expansion of cube `=8xx10^(-6).^(@)C^(-1)`. The distance of bottom of cube from plate does not change during warming up. Choose the correct option (s)

Options

A. The change in temperature of cube so that it would touch the plate is `6.2^(@)C`
B. The change in temperature of cube so that it would touch the plate is `3.1^(@)`
C. The second wavelength, which gives maximum reinforcement is `lamda=2000/3nm`
D. The second wavelength, which gives maximum reinforce is `lamda=4000/3nm`

Answer

B, C

Explanation

Condition for maximum intensity due to thin film
`2mud-(lamda)/2=nlamda`
`implies2mud=(2p+1)(lamda_(1))/2implies(2p+1)lamda_(1)=4mud`
Here `lamda_(1)=400nm` and `n=pimplies` An integer
Let `lamda^(')=1150nm`. Since only two wavelength give maximum intensity, so
`(4mud)/(2(p-1)+1) lt lamda^(') lt (4mud)/(2(p-2)+1)`
`(lamda_(1)(2p+1))/(2(p-1)+1)lt lamda^(')(lamda_(1)(2p+1))/(2(p-2)+1)`
`implies(lamda_(1)(2p+1))/(2(p-1)+1) lt lamda^(')` or `lamda_(') lt (lamda_(1)(2p+1))/(2(p-2)+1)`
`implies p gt 1/2 ((lamda^(')+lamda_(1))/(lamda^(')+lamda_(1)))=1/2((1150+400)/(1150-400))=1..........` or
`p lt 1/2 ((3lamda^(')+lamda_(1))/(lamda^(')-lamda_(1)))=1/2((3xx1150+400)/(1150-400))=2`.........
The only integer which satisfies both inequalities is 2 so
`d=((2xx2+1)400xx10^(-9))/(4xx1)=500nm`
`lamda_(2)=(4mud)/(2(p-2)+1)=(4xx1xx500xx10^(-9))/(2+1)=666.7nm`
`Deltat=d/(alphah)=(500xx10^(-9))/(8xx10^(-6)xx2xx10^(-2))~~3.1^(@)C`