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In a Young's double slit experiment, the...

In a Young's double slit experiment, the angular separation of the interference fringes on a distance screen is `0.04A^(0)`. The angular separation if the entire apparatus is immersed in a liquid of R.I. `4/3` is `0.0nA^(0)`, then what is the value of `n`?

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To solve the problem, we need to understand the relationship between the angular separation of the interference fringes in a Young's double slit experiment and the refractive index of the medium in which the apparatus is placed. ### Step-by-Step Solution: 1. **Identify the Given Values**: - Angular separation in air (α_A) = 0.04 A°. - Refractive index of the liquid (μ) = 4/3. - Angular separation in the liquid (α_L) = 0.0 n A°. 2. **Understand the Relationship**: The angular separation (α) is directly proportional to the wavelength (λ) of the light used. When the apparatus is immersed in a medium with a refractive index, the wavelength of light changes according to the formula: \[ \lambda_L = \frac{\lambda_A}{\mu} \] where: - λ_A = wavelength in air, - λ_L = wavelength in the liquid, - μ = refractive index of the liquid. 3. **Relate Angular Separations**: The angular separations in air and liquid can be related as follows: \[ \frac{\alpha_A}{\alpha_L} = \frac{\lambda_A}{\lambda_L} \] Substituting the expression for λ_L: \[ \frac{\alpha_A}{\alpha_L} = \frac{\lambda_A}{\frac{\lambda_A}{\mu}} = \mu \] 4. **Substituting Known Values**: Now substituting the known values into the equation: \[ \frac{0.04}{0.0 n} = \frac{4}{3} \] 5. **Cross-Multiplying**: Cross-multiplying gives: \[ 0.04 \cdot 3 = 0.0 n \cdot 4 \] Simplifying this: \[ 0.12 = 0.0 n \cdot 4 \] 6. **Solving for n**: Dividing both sides by 4: \[ n = \frac{0.12}{4} = 0.03 \] 7. **Final Calculation**: Since we have \(0.0 n\) in the angular separation, we can express n as: \[ n = \frac{0.12}{4} = 0.03 \text{ (but we need to express it in terms of a whole number)} \] To express it in terms of a whole number, we can multiply both sides by 100: \[ n = \frac{12}{4} = 3 \] ### Conclusion: Thus, the value of \(n\) is **3**.

To solve the problem, we need to understand the relationship between the angular separation of the interference fringes in a Young's double slit experiment and the refractive index of the medium in which the apparatus is placed. ### Step-by-Step Solution: 1. **Identify the Given Values**: - Angular separation in air (α_A) = 0.04 A°. - Refractive index of the liquid (μ) = 4/3. - Angular separation in the liquid (α_L) = 0.0 n A°. ...
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