Question:medium

In Young's double slit experiment, the distance between screen and aperture is $1\text{ m}$ . The slit width is $2\text{ mm}$ . Light of $6000\text{ \AA}$ is used. If a thin glass plate ( $\mu = 1.5$ ) of thickness $0.04\text{ mm}$ is placed over one of the slits, then there will be a lateral displacement of the fringes by}

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Lateral shift $\Delta x = \beta \frac{(\mu - 1)t}{\lambda}$, where $\beta$ is fringe width. The pattern always shifts toward the side where the glass plate is placed.
Updated On: May 14, 2026
  • 0.5 cm
  • 1 cm
  • 1.5 cm
  • 2 cm
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
When a transparent medium (like a glass plate) of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the interfering beams in Young's Double Slit Experiment (YDSE), the entire fringe pattern shifts laterally.
Step 2: Key Formula or Approach:
The introduction of the slab creates an additional optical path difference of $(\mu - 1)t$.
The formula for the lateral shift (displacement) $y_0$ of the central fringe (and thus the whole pattern) is:
\[ y_0 = \frac{D}{d} (\mu - 1)t \]
where $D$ is the distance to the screen, $d$ is the slit separation (often referred to as slit width in this context), $\mu$ is the refractive index, and $t$ is the thickness of the plate.
Step 3: Detailed Explanation:
Given values:
Distance to screen, $D = 1\text{ m}$
Slit separation, $d = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$
Refractive index, $\mu = 1.5$
Thickness of plate, $t = 0.04\text{ mm} = 0.04 \times 10^{-3}\text{ m}$
Now, substitute these values into the displacement formula:
\[ y_0 = \frac{1}{2 \times 10^{-3}} (1.5 - 1) \times (0.04 \times 10^{-3}) \]
\[ y_0 = \frac{1}{2 \times 10^{-3}} \times 0.5 \times 0.04 \times 10^{-3} \]
The $10^{-3}$ terms cancel out:
\[ y_0 = \frac{0.5 \times 0.04}{2} \]
\[ y_0 = \frac{0.02}{2} = 0.01\text{ m} \]
Convert the result to centimeters to match the options:
\[ y_0 = 0.01\text{ m} = 1\text{ cm} \]
Step 4: Final Answer:
The lateral displacement of the fringes is $1\text{ cm}$.
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