Question

When waves from two coherent sources, having amplitudes a and b superimpose, the amplitude R of the resultant wave is given by : 

= \(\sqrt{a^2+b^2+2ab\,cosϕ}\), 

Where φ is the constant phase angle between the two waves. 

The resultant intensity I is directly proportional to the square of the amplitude of the resultant wave, 

I.e. I∝(R²) 

I.e. I ∝ (a² + b² + 2ab cosϕ) 

For constructive interference,

φ = 2nπ, 

I_max = (a + b)² 

For destructive interference,

φ = (2n-1)π 

I_min = (a − b)² 

If, I₁ and I₂ are intensities of light from two slits of widths, w₁ and w₂, 

Then,

\(\frac{I_1}{I_2}\) = \(\frac{w_1}{w_2}\) = \(\frac{a^2}{b^2}\) 

Light waves from two coherent sources of intensity ratio 81:1 produce interference.

(i) The ratio of amplitudes of two sources is : 

(a) 9:1 

(b) 81:1 

(c) 1:9 

(d) 1:81 

(ii) The ratio of slit widths of the two sources is : 

(a) 9:1 

(b) 81:1 

(c) 1:9 

(d) 1:81

(iii) The ratio of maxima and minima in the interference pattern is : 

(a) 9:1 

(b) 81:1 

(c) 25:16 

(d) 16:25 

(iv) If two slits in Young’s experiment have width ratio 1:4, the ratio of maximum and minimum intensity in the interference pattern would be : 

(a) 1:4 

(b)1:16 

(c) 9:1 

(d) 9:16

Answer 1

(i). Option : (a) 

Here,

\(\frac{I_1}{I_2}\) = \(\frac{81}{1}\) 

If a, b are respective amplitudes, 

Then,

\(\frac{I_1}{I_2}\) = \(\frac{a^2}{b^2}\)= \(\frac{81}{1}\) 

⇒ \(\frac{a}{b}\) = \(\frac{9}{1}\)

(ii). Option : (b) 

If W₁ and W₂ are the slit widths, then

\(\frac{W_1}{W_2}\) = \(\frac{I_1}{I_2}\) = \(\frac{81}{1}\)

(iii). Option : (c) 

We have,

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

As, 

a = 9b 

[from question (i)]

\(\frac{I_{max}}{I_{min}}\) = \(\frac{(9b+b)^2}{(9b-b)^2}\) 

= \(\frac{(10b)^2}{(8b)^2}\) 

= \(\frac{25}{16}\)

(iv). Option : (c) 

Here,

\(\frac{W_1}{W_2}\) = \(\frac{1}{4}\) 

= \(\frac{a^2}{b^2}\)