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The ratio of the intensities at minima t...

The ratio of the intensities at minima to maxima in the interference pattern is 9 : 25. What will be the ratio of the widths of the two slits in the young's double slit experiment ?

A

`8 : 1`

B

`16 : 1`

C

`4 : 1`

D

`9 : 1`

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AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the ratio of the widths of the two slits in Young's double slit experiment based on the given ratio of intensities at minima to maxima. ### Step-by-Step Solution: 1. **Understand the Given Ratio**: The problem states that the ratio of the intensities at minima to maxima is given as 9:25. This means: \[ I_{min} : I_{max} = 9 : 25 \] 2. **Express Intensities in Terms of Slit Widths**: In Young's double slit experiment, the intensity at maxima \(I_{max}\) and minima \(I_{min}\) can be expressed in terms of the amplitudes (or widths) of the two slits \(a_1\) and \(a_2\): \[ I_{max} = (a_1 + a_2)^2 \] \[ I_{min} = (a_1 - a_2)^2 \] 3. **Set Up the Ratio of Intensities**: From the given ratio, we can write: \[ \frac{I_{min}}{I_{max}} = \frac{9}{25} \] Substituting the expressions for \(I_{min}\) and \(I_{max}\): \[ \frac{(a_1 - a_2)^2}{(a_1 + a_2)^2} = \frac{9}{25} \] 4. **Cross Multiply to Eliminate the Fraction**: Cross multiplying gives us: \[ 25(a_1 - a_2)^2 = 9(a_1 + a_2)^2 \] 5. **Expand Both Sides**: Expanding both sides: \[ 25(a_1^2 - 2a_1a_2 + a_2^2) = 9(a_1^2 + 2a_1a_2 + a_2^2) \] This simplifies to: \[ 25a_1^2 - 50a_1a_2 + 25a_2^2 = 9a_1^2 + 18a_1a_2 + 9a_2^2 \] 6. **Rearranging the Equation**: Rearranging gives: \[ 25a_1^2 - 9a_1^2 - 50a_1a_2 - 18a_1a_2 + 25a_2^2 - 9a_2^2 = 0 \] Which simplifies to: \[ 16a_1^2 - 68a_1a_2 + 16a_2^2 = 0 \] 7. **Factoring the Quadratic**: Dividing through by 16: \[ a_1^2 - 4.25a_1a_2 + a_2^2 = 0 \] This can be factored or solved using the quadratic formula. 8. **Finding the Ratio of Slit Widths**: By solving the quadratic equation, we find: \[ a_1 = 4a_2 \] Therefore, the ratio of the widths of the two slits is: \[ \frac{a_1}{a_2} = 4:1 \] ### Final Answer: The ratio of the widths of the two slits is \(4:1\). ---
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AAKASH SERIES-WAVES OPTICS-EXERCISE -III (DOPPLER EFFECT IN LIGHT, INTERFERENCE)
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  2. Light from two coherent sources of same amplitude and same wavelength ...

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  3. In Young's double slit experiment an interference pattern is obtained ...

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  4. In young's double slit experiment the n^(th) red bright band coincides...

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  5. In Young's double slit experiment, 12 fringes are observed to be forme...

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  6. In Young.s double slit experiment the distance between the sources is ...

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  7. In Young's experiment interference bands are produced on the screen pl...

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  8. The two coherent sources of equal intensity produce maximum intensity ...

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  9. The ratio of the intensities at minima to maxima in the interference p...

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  10. Two coherent monochromatic light sources are located at two vertices o...

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  11. In Young's double slit experiment S(1) and S(2) are two slits. Films o...

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  12. In a Young's double slit experiment using monochromatic light, the fri...

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  13. When a thin transparent plate of Refractive Index 1.5 is introduced in...

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  14. Two coherent point sources S1 " and "S2 vibrating in phase light of wa...

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  15. A transparent glass plate of thickness 0.5 mm and refractive index 1.5...

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  16. In the Young's double slit experiment using a monochromatic light of w...

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  17. YDSE is carried with two thin sheets of thickness 10.4mu m each and r...

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  18. In the Young's double slit experiment, the intensity of light at a poi...

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  19. In Young.s double slit experiment, the slits are 2mm apart and are ill...

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  20. In a Young.s interference experimental arrangement, the yellow light i...

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