The ratio of maximum to minimum intensity due to superposition of two waves is 949. Then the ratio of the intensity of component waves is

We know that, a∝I2So, IminImax=949So, aminamax=37So, aminamax=a1−a2a1+a2=7−37+3∴a2a1=25As, a∝I2∴I2I1=425

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The ratio of maximum to minimum intensity due to superposition

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The ratio of maximum to minimum intensity due to superposition of two waves is $\frac{49}{9}$ . Then the ratio of the intensity of component waves is

\frac{25}{4}

\frac{16}{25}

\frac{4}{49}

\frac{9}{49}

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Oliver Discussed

The ratio of maximum to minimum intensity due to superposition of two waves is

\frac{49}{9}

. Then the ratio of the intensity of component waves is

12 mins ago

Discuss this question LIVE

12 mins ago

We know that

, a \propto I^{2}

So,

\frac{I _{ma x}}{I _{min}} = \frac{49}{9}

So,

\frac{a _{ma x}}{a _{min}} = \frac{7}{3}

So,

\frac{a _{ma x}}{a _{min}} = \frac{a _{1} + a _{2}}{a _{1} - a _{2}} = \frac{7 + 3}{7 - 3}

∴ \frac{a _{1}}{a _{2}} = \frac{5}{2}

As,

a \propto I^{2}

∴ \frac{I _{1}}{I _{2}} = \frac{25}{4}

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Question Text	The ratio of maximum to minimum intensity due to superposition of two waves is $\frac{49}{9}$ . Then the ratio of the intensity of component waves is
Answer Type	Text solution:1
Upvotes	150