IIT JEE , AIEEE Forum - Physics , Optics

joyfrancis

q) If the ratio of the intensities of two interferring light waves are 4:1 then the ratio of the intensities of the maxima and the minima are?

the answer given is 9:1 , but im getting 5:3

by phasor's method it is clear that the intensity of the new wave formed is given by

sqroot(16I^2 + I^2 + 8I^2cos@)..where @ is the phase difference b/w the two waves

Imax is when cos@=1 so Imax = 5I

and Imin is when cos@=-1 so Imin = 3I

so Imax / Imin = 5/3

ramkumar_november

given

$\frac{I_1}{I_2}=\frac{4}{1}$

$\frac{I_1}{I_2}=\frac{a^2}{b^2}$ (a and b are amplitudes of the interfering light waves )

$\frac{a^2}{b^2}=\frac{4}{1}$

a=2b

$\frac{I_{max}}{I_{min}}=\frac{(a+b)^2}{(a-b)^2}$

$\frac{I_{max}}{I_{min}}=\frac{(2b+b)^2}{(2b-b)^2}$

$\frac{I_{max}}{I_{min}}=\frac{9}{1}$

rhd92781

In such type of problems, simply do

I₁/I₂ = 4/1

$\sqrt{I1}/\sqrt{I2} = 2/1$

$(\sqrt{I1}+\sqrt{I2})/(\sqrt{I1}-\sqrt{I2}) = (2+1)/(2-1) = 3:1$

Now I_max: I_min = $(\sqrt{I1} + \sqrt{I2})^{2} : (\sqrt{I1}-\sqrt{I2})^{2}$ = 9:1