Question:
Interference fringes produced by a double slit arrangement using a monochromatic light of wave length 5,890Å have an angular fringe width \( 0.28^\circ \). If the entire arrangement is immersed in water, the new angular fringe width will be (\( ^a\mu_w = 4/3 \)):
Interference fringes produced by a double slit arrangement using a monochromatic light of wave length 5,890Å have an angular fringe width \( 0.28^\circ \). If the entire arrangement is immersed in water, the new angular fringe width will be (\( ^a\mu_w = 4/3 \)):
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Immersing a YDSE setup in a medium increases the effective refractive index, causing the fringe width to decrease by a factor of \( \mu \).
Updated On: Jun 9, 2026
- \( 0.24^\circ \)
- \( 0.21^\circ \)
- \( 0.18^\circ \)
- \( 0.36^\circ \)
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The Correct Option is B
Solution and Explanation
Concept:
The angular fringe width in a double-slit experiment is given by \( \beta_\theta = \frac{\lambda}{d} \), where \(\lambda\) is the wavelength and \(d\) is the slit separation. When immersed in a medium of refractive index \( \mu \), the wavelength changes to \( \lambda' = \frac{\lambda}{\mu} \).
Step 1: Relate fringe width to refractive index.
Initially, \( \beta_\theta = \frac{\lambda}{d} \). In the medium (water), the new wavelength is \( \lambda' = \frac{\lambda}{\mu} \), so the new angular fringe width \( \beta_\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} \).
Step 2: Establish the ratio of fringe widths.
$$ \frac{\beta_\theta'}{\beta_\theta} = \frac{1}{\mu} $$ $$ \beta_\theta' = \frac{\beta_\theta}{\mu} $$
Step 3: Calculate the new width.
Given \( \beta_\theta = 0.28^\circ \) and \( \mu = 4/3 \): $$ \beta_\theta' = \frac{0.28}{4/3} = 0.28 \times \frac{3}{4} $$ $$ \beta_\theta' = 0.07 \times 3 = 0.21^\circ $$ $$\boxed{0.21^\circ}$$
Step 1: Relate fringe width to refractive index.
Initially, \( \beta_\theta = \frac{\lambda}{d} \). In the medium (water), the new wavelength is \( \lambda' = \frac{\lambda}{\mu} \), so the new angular fringe width \( \beta_\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} \).
Step 2: Establish the ratio of fringe widths.
$$ \frac{\beta_\theta'}{\beta_\theta} = \frac{1}{\mu} $$ $$ \beta_\theta' = \frac{\beta_\theta}{\mu} $$
Step 3: Calculate the new width.
Given \( \beta_\theta = 0.28^\circ \) and \( \mu = 4/3 \): $$ \beta_\theta' = \frac{0.28}{4/3} = 0.28 \times \frac{3}{4} $$ $$ \beta_\theta' = 0.07 \times 3 = 0.21^\circ $$ $$\boxed{0.21^\circ}$$
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