Question

The equation of two light waves are $y_1=6cos\omega t,y_2=8cos(\omega t+\varphi )$ the ratio of maximum to minimum intensities produced by the superposition of these wave will be

• A. 49 :1
• B. 1 : 49
• C. 1 : 7
• D. 7: 1

Answer 1

Answer: (A)

Step 1: Given that:

Equation of light waves;

$y_1=6\cos \text{ω}t$

$y_2=8\cos (\text{ω}t+\text{φ})$

Step 2: Concept and formula used:

When two waves are superposed,

The maximum amplitude of the resultant wave is the sum of the amplitude of the individual waves and the minimum amplitude of the resultant wave is the difference of the amplitude of the individual waves.

Thus,

If $y_1=A_1\cos \text{ω}t$ and $y_2=A_2\cos (\text{ω}t+\text{φ})$ be two waves superposing on each other,

Then the maximum amplitude of the resulting wave = $A_1+A_2$

The minimum amplitude of the resulting wave = $A_1-A_2$

The intensity of a wave $\propto$ square of the amplitude of the wave

That is

$I\propto A^2$

$I=k{A}^2$

Step 3: calculation of the ratio of the maximum and minimum intensities of the resulting wave:

Maximum amplitude($A_{max}$) = $6unit+8unit=14unit$

Minimum amplitude($A_{min}$) = $6unit-8unit=-2unit$

Thus,

Maximum intensity($I_{max}$) = $k{\left(14unit\right)}^2=196kuni{t}^2$

Minimum intensity($I_{min}$) = $k{(-2unit)}^2=4kuni{t}^2$

Thus,

$\frac{I_{max}}{I_{min}}=\frac{196kuni{t}^2}{4kuni{t}^2}$

$\frac{I_{max}}{I_{min}}=\frac{49}{1}$

Thus,

The ratio of maximum to minimum intensities produced by the superposition of the given wave will be $49:1$ .