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The two coherent sources of equal intens...

The two coherent sources of equal intensity produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by 50% by reducing its width then the intensity of light at the same point will be

A

90

B

89

C

67

D

`72.85`

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The correct Answer is:
To solve the problem, we need to calculate the resultant intensity when one of the coherent sources has its intensity reduced by 50%. Let's break down the solution step by step. ### Step-by-Step Solution: 1. **Understanding the Initial Conditions**: - We have two coherent sources of equal intensity, producing a maximum intensity of 100 units at a point. - Let the intensity of each source be \( I_0 \). 2. **Finding the Initial Intensity**: - The maximum intensity \( I_{\text{max}} \) for two coherent sources is given by: \[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \theta \] - For maximum intensity, \( \cos \theta = 1 \): \[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \] - Since both sources are equal, \( I_1 = I_2 = I_0 \): \[ I_{\text{max}} = 2I_0 + 2\sqrt{I_0 I_0} = 2I_0 + 2I_0 = 4I_0 \] - Given \( I_{\text{max}} = 100 \): \[ 4I_0 = 100 \implies I_0 = 25 \text{ units} \] 3. **Adjusting the Intensity of One Source**: - One of the sources has its intensity reduced by 50%, so the new intensities are: - \( I_1 = I_0 = 25 \) units - \( I_2 = \frac{I_0}{2} = \frac{25}{2} = 12.5 \) units 4. **Calculating the New Resultant Intensity**: - Using the formula for maximum intensity again: \[ I_{\text{new}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \] - Substituting the new values: \[ I_{\text{new}} = 25 + 12.5 + 2\sqrt{25 \times 12.5} \] - Calculate \( \sqrt{25 \times 12.5} \): \[ \sqrt{25 \times 12.5} = \sqrt{312.5} \approx 17.68 \] - Now substituting back: \[ I_{\text{new}} = 25 + 12.5 + 2 \times 17.68 \approx 25 + 12.5 + 35.36 \] - Adding these values: \[ I_{\text{new}} \approx 25 + 12.5 + 35.36 = 72.86 \text{ units} \] 5. **Final Result**: - The intensity of light at the same point after reducing the intensity of one source by 50% is approximately **72.85 units**.
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