The equation of two light waves are y1-6 cos- t- y2-8 cos- t- - the ratio of maximum to minimum intensities produced by the superposition of these wave will be

Y_1=6 cos⁡ ω t ; y_2=8 cos⁡(ω t+ ϕ ) I_max/I_min = ( 8+6/8 6 )^2= 49/1 ...

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Standard XII

Physics

Intensity Explanation in YDSE

The equation ...

Question

The equation of two light waves are $y_{1} = 6 c o s ω t, y_{2} = 8 c o s (ω t + ϕ)$ the ratio of maximum to minimum intensities produced by the superposition of these wave will be

49 :1

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1 : 49

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1 : 7

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Solution

The correct option is A 49 :1
$y_{1} = 6 c o s ω t$ ; $y_{2} = 8 c o s (ω t + ϕ)$
$\frac{I_{m a x}}{I_{m i n}} = {(\frac{8 + 6}{8 - 6})}^{2} = \frac{49}{1}$

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The equation of two light waves are y1=6cosωt,y2=8cos(ωt+ϕ) the ratio of maximum to minimum intensities produced by the superposition of these wave will be

The correct option is A 49 :1y1=6cosωt ; y2=8cos(ωt+ϕ) ImaxImin=(8+68−6)2=491

The equation of two light waves are $y_{1} = 6 c o s ω t, y_{2} = 8 c o s (ω t + ϕ)$ the ratio of maximum to minimum intensities produced by the superposition of these wave will be

The correct option is A 49 :1
$y_{1} = 6 c o s ω t$ ; $y_{2} = 8 c o s (ω t + ϕ)$
$\frac{I_{m a x}}{I_{m i n}} = {(\frac{8 + 6}{8 - 6})}^{2} = \frac{49}{1}$