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Standard XII

Physics

Interference General Case

If the nbs...

Question

If the intensity ratio of two coherent sources used in Young's double slit experiment is $49 : 1$ , then the ratio between the maximum and minimum intensities in the interference pattern is

1 : 9

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9 : 16

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25 : 16

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16 : 25

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16 : 9

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Solution

The correct option is C $16 : 9$
The intensity ratio of two coherent sources becomes
$I_{1} = 49 \times I_{2}$
$I_{m a x} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = 64 I_{2}$
$I_{m i n} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}} = 36 I_{2}$
$\frac{I_{m a x}}{I_{m i n}} = \frac{64}{36} = 16 : 9$

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If the intensity ratio of two coherent sources used in Young's double slit experiment is $49 : 1$ , then the ratio between the maximum and minimum intensities in the interference pattern is

The correct option is C $16 : 9$
The intensity ratio of two coherent sources becomes
$I_{1} = 49 \times I_{2}$
$I_{m a x} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = 64 I_{2}$
$I_{m i n} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}} = 36 I_{2}$
$\frac{I_{m a x}}{I_{m i n}} = \frac{64}{36} = 16 : 9$

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If the intensity ratio of two coherent sources used in Young's double slit experiment is 49:1, then the ratio between the maximum and minimum intensities in the interference pattern is

The correct option is C 16:9The intensity ratio of two coherent sources becomesI1=49×I2Imax=I1+I2+2√I1I2=64I2Imin=I1+I2−2√I1I2=36I2ImaxImin=6436=16:9

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STATE BOARDS

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If the intensity ratio of two coherent sources used in Young's double slit experiment is $49 : 1$ , then the ratio between the maximum and minimum intensities in the interference pattern is

The correct option is C $16 : 9$
The intensity ratio of two coherent sources becomes
$I_{1} = 49 \times I_{2}$
$I_{m a x} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = 64 I_{2}$
$I_{m i n} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}} = 36 I_{2}$
$\frac{I_{m a x}}{I_{m i n}} = \frac{64}{36} = 16 : 9$