You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Intensity Amplitude Relationship

Two waves of ...

Question

Two waves of intensity $l_{2} and I_{2}$ interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, $I_{1} / I_{2}$ is equal to

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C

$\frac{I_{m a x}}{I_{m i n}} = \frac{(\sqrt{I_{1}} + \sqrt{I_{2}})^{2}}{(\sqrt{I_{1}} - \sqrt{I_{2}})^{2}} = a \frac{I_{1}}{I_{2}} = {[\frac{\sqrt{a} + 1}{\sqrt{a} + 1}]}^{2} i f I_{1} > I_{2}$

Suggest Corrections

Similar questions

Q. Two waves of intensity

I_{1}

and

I_{2}

interfere in a medium. If the ratio of maximum intensity produced to the minimum intensity is

α

, then

\frac{I_{1}}{I_{2}}

is equal to

Two waves of intensity $I_{2}$ and $I_{2}$ interfere. The maximum intensity produced to the minimum intensity is in the ratio $α$ . Then, $\frac{I_{1}}{I_{2}}$ is equal to

Q. Two beams of light of intensity

I_{1}

and

I_{2}

interfere to give an interference pattern. If the ratio of maximum intensity to that of minimum intensity is

\frac{25}{9}

, then ratio of

\frac{I_{1}}{I_{2}}

Q. In a photoelectric experiment, intensity of incident light is varied and the following graph between photo current and potential is plotted. Find the relation between

I_{1}

and

I_{2}

Let $I_{1} = {(\frac{π}{4})}^{2} + \sqrt{2}, I_{2} = {(t a n^{- 1} (\frac{1}{e}))}^{2} + \frac{2 e}{\sqrt{e^{2} + 1}}, I_{3} = (t a n^{- 1} e)^{2} + \frac{2}{\sqrt{e^{2} + 1}}$ , then which of the following is true?

Join BYJU'S Learning Program

Two waves of intensity l2 and I2 interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, I1/I2is equal to

The correct option is C ImaxImin=(√I1+√I2)2(√I1−√I2)2=aI1I2=[√a+1√a+1]2 if I1>I2

Two waves of intensity $l_{2} and I_{2}$ interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, $I_{1} / I_{2}$ is equal to

The correct option is C

$\frac{I_{m a x}}{I_{m i n}} = \frac{(\sqrt{I_{1}} + \sqrt{I_{2}})^{2}}{(\sqrt{I_{1}} - \sqrt{I_{2}})^{2}} = a \frac{I_{1}}{I_{2}} = {[\frac{\sqrt{a} + 1}{\sqrt{a} + 1}]}^{2} i f I_{1} > I_{2}$