Two coherent sources whose intensity ratio is 64: 1 produce interference fringes. The ratio of intensities of maxima and minima is

Question

Two coherent sources whose intensity ratio is 64: 1 produce interference fringes. The ratio of intensities of maxima and minima is (A) 9: 7 (B) 8: 1 (C) 81: 49 (D) 81: 7

Tardigrade Admin · Accepted Answer

The ratio of intensities of maxima and minima (I max /I min )=((√I1+√I2)2/(√I1-√I2)2)=((√64+√1)2/(√64-√1)2) (I max /I min )=((8+1)2/(8-1)2)=(81/49)

Q. Two coherent sources whose intensity ratio is 64 : 1 produce interference fringes. The ratio of intensities of maxima and minima is

9: 7

8 : 1

81 : 49

81 : 7

The ratio of intensities of maxima and minima $\frac{I _{m a x}}{I _{m i n}} = \frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = \frac{( 64 + 1 ) ^{2}}{( 64 - 1 ) ^{2}}$ $\frac{I _{m a x}}{I _{m i n}} = \frac{( 8 + 1 ) ^{2}}{( 8 - 1 ) ^{2}} = \frac{81}{49}$

Q. Two coherent sources whose intensity ratio is 64 : 1 produce interference fringes. The ratio of intensities of maxima and minima is

9: 7

8 : 1

81 : 49

81 : 7

The ratio of intensities of maxima and minima Imin​Imax​​=(I1​​−I2​​)2(I1​​+I2​​)2​=(64​−1​)2(64​+1​)2​ Imin​Imax​​=(8−1)2(8+1)2​=4981​

The ratio of intensities of maxima and minima $\frac{I _{m a x}}{I _{m i n}} = \frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = \frac{( 64 + 1 ) ^{2}}{( 64 - 1 ) ^{2}}$ $\frac{I _{m a x}}{I _{m i n}} = \frac{( 8 + 1 ) ^{2}}{( 8 - 1 ) ^{2}} = \frac{81}{49}$