3Δx1=4/cos(0.01) - (4/cos0.01)/sin(pie/2 - 0.02)
Δx2=4/cos(0.03) - (4/cos0.03)/sin(pie/2 - 0.06)
Δx1 - Δx2 = wavelength = the answer
22 Answers
3
21NOPES DA ANS IS A........
I ASKD U 2 POST UR WORKIN SO V CAN FIND DA MISTAKE IF NE.......
21LOL..... SORRY MISS∫∫ DINNO.... HEHE........
NEWAYS SUM1 POST DA CALCULATN PART PL... OR FIND A MISTAKE IN MY DIAG. [6]
39lolzzzzzzzzzzzzzzzz
abhishek aaj kar pad kar kuch yaad nahiin rah raha hain
21nihit's soln is almost perfect........
1 and 2 written in diag are 0.01 & 0.02...... lol.....
if u get da precise ans frm here on... pl post da soln.....
33sayad ho gaya...
Δx1=4/cos(0.01)
4/cos(0.01)=nλ
Δx2=4/cos(0.03)
4/cos(0.03)=nλ+λ
4/cos(0.03)=4/cos(0.01)+λ
λ=4/cos(0.03)-4/cos(0.01)
but from it.. λ=0.0016 m [2]
21the crct answers are .15 for Q1
.16 for Q2 as u can c.....
it looks like an interfrnce que so the diff in path length mite b vital!!!!
jus guessing...........
33hey... i am getting.... veryyyyyyyyyy wrong answer... 800/3 metres
:D
3optically denser i think
but dont know wat is the ratio of the waves dat got reflected
33[3] i have gone mad... it is rarer.. so no phase diff due to that...
33after reflection from water... will there be a phase diff due to reflection.. water is rarer or denser...for radio waves..
1man its difficult and different...........im tryin all sorts of methods...