Question:medium

The width of the fringes obtained with light of wavelength $2 \times 10^{-7}\text{ m}$ is $82\text{ mm}$. If the entire apparatus is immersed in a liquid of refractive index $3$, what will be the width of the resulting fringes?

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The initial wavelength value ($6.2 \times 10^{-7}\text{ m}$) is extra information provided to look complicated. Since fringe width scales inversely with the refractive index ($\beta' = \frac{\beta}{\mu}$), you only need to divide the initial width by the index value to get the answer.
Updated On: May 20, 2026
  • $1.82\text{ mm}$
  • $0.71\text{ mm}$
  • $2.8\text{ mm}$
  • $1.4\text{ mm}$
Show Solution

The Correct Option is D

Solution and Explanation

Understanding the Concept: In a Young's Double Slit Experiment setup, the linear width of the interference fringes ($\beta$) is given by: \[ \beta = \frac{\lambda D}{d} \implies \beta \propto \lambda \] When the entire experimental physical apparatus is submerged in a liquid medium of refractive index $\mu$, the physical speed of light drops, causing its operational wavelength to shorten to $\lambda' = \frac{\lambda}{\mu}$. Consequently, the fringe width decreases by the same index factor: \[ \beta' = \frac{\beta}{\mu} \]
Step 1: Substitute values directly into the scaling expression.
We are given:
Initial fringe width, $\beta = 1.82\text{ mm}$
Refractive index of the liquid, $\mu = 1.3$
Let's compute the modified fringe width ($\beta'$): \[ \beta' = \frac{1.82\text{ mm}}{1.3} = 1.4\text{ mm} \]
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