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Two monochromatic light beams of intensi...

Two monochromatic light beams of intensity 16 and 9 units are interfering. The ratio of inetnsities of bright and dark parts of the resultant pattern is :

A

`16/9`

B

`4/3`

C

`7/1`

D

`49/1`

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The correct Answer is:
To solve the problem of finding the ratio of intensities of bright and dark parts of the resultant pattern formed by two interfering monochromatic light beams with intensities \( I_1 = 16 \) units and \( I_2 = 9 \) units, we will follow these steps: ### Step-by-Step Solution: 1. **Identify the Intensities**: - Let \( I_1 = 16 \) units and \( I_2 = 9 \) units. 2. **Use the Formulas for Maximum and Minimum Intensities**: - The formula for maximum intensity \( I_{max} \) is given by: \[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \] - The formula for minimum intensity \( I_{min} \) is given by: \[ I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] 3. **Calculate \( \sqrt{I_1} \) and \( \sqrt{I_2} \)**: - Calculate \( \sqrt{I_1} = \sqrt{16} = 4 \) - Calculate \( \sqrt{I_2} = \sqrt{9} = 3 \) 4. **Substitute into the Maximum Intensity Formula**: - Substitute the values into the \( I_{max} \) formula: \[ I_{max} = (4 + 3)^2 = 7^2 = 49 \] 5. **Substitute into the Minimum Intensity Formula**: - Substitute the values into the \( I_{min} \) formula: \[ I_{min} = (4 - 3)^2 = 1^2 = 1 \] 6. **Find the Ratio of Intensities**: - The ratio of maximum intensity to minimum intensity is: \[ \frac{I_{max}}{I_{min}} = \frac{49}{1} = 49 \] ### Final Answer: The ratio of the intensities of the bright and dark parts of the resultant pattern is \( 49:1 \).

To solve the problem of finding the ratio of intensities of bright and dark parts of the resultant pattern formed by two interfering monochromatic light beams with intensities \( I_1 = 16 \) units and \( I_2 = 9 \) units, we will follow these steps: ### Step-by-Step Solution: 1. **Identify the Intensities**: - Let \( I_1 = 16 \) units and \( I_2 = 9 \) units. 2. **Use the Formulas for Maximum and Minimum Intensities**: ...
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DISHA PUBLICATION-WAVE OPTICS-EXERCISE-2 : CONCEPT APPLICATOR
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  3. In a two slit experiment with monochromatic light fringes are obtained...

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  4. In Young's experiment the distance between two slits is d/3 and the di...

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  8. The central fringe of the interference pattern produced by the light o...

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  9. Figure shows two light rays that are initially exactly in phase and th...

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  10. In young's double slit experiment the slits are illumated by light of ...

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  11. In a double slit experiment, the two slits are 1 mm apart and the scre...

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  12. In a Young's double slit experiment, the two slits act as coherent sou...

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  13. In Young's double slit experiment, the fringes are displaced index 1.5...

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  14. There are two plane mirrors. They are mutually inclined as shown in fi...

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  15. In the ideal double-slit experiment, when a glass-plate(refractive ind...

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  16. A thin glass plate of thickness is (2500)/3lamda (lamda is wavelength ...

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  17. In Young's double slit experiment, wavelength lambda=5000Å the distanc...

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  18. A person lives in a high-rise building on the bank of a river 50 m wid...

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  19. A broad sources of light of wavelength 680nm illuminated normally two ...

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  20. A thin film of soap solution (mu(s)=1.4) lies on the top of a glass pl...

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