Question

Young's double slit experiment is first done in air and then in a medium of refractive index \(\mu\). If the \(7\)th dark fringe in the medium lies where the \(4\)th bright fringe is in air, then the value of \(\mu\) is:

• A. \(1.654\)
• B. \(1.389\)
• C. \(1.875\)
• D. \(1.768\)

Hint:

In a medium, fringe width becomes \(\beta' = \frac{\beta}{\mu}\). For dark fringes, carefully check the counting convention used in the question.

Answer 1

The Correct Option is C

Step 1: Write position of bright fringe in air.
In Young's double slit experiment, position of \(n\)th bright fringe is:
\[ y_n = n\beta \]
For \(4\)th bright fringe in air:
\[ y = 4\beta \]

Step 2: Write fringe width in air.
\[ \beta = \frac{\lambda D}{d} \]

Step 3: Write fringe width in medium.
In a medium of refractive index \(\mu\), wavelength becomes:
\[ \lambda' = \frac{\lambda}{\mu} \]
So, fringe width in medium becomes:
\[ \beta' = \frac{\beta}{\mu} \]

Step 4: Write position of dark fringe in medium.
Position of \(n\)th dark fringe is:
\[ y_n = \left(n - \frac{1}{2}\right)\beta' \]
For \(7\)th dark fringe in medium:
\[ y = \left(7 - \frac{1}{2}\right)\beta' \]
\[ y = \frac{13}{2}\beta' \]

Step 5: Equate given positions.
According to the question, \(7\)th dark fringe in medium lies where \(4\)th bright fringe is in air.
\[ \frac{13}{2}\beta' = 4\beta \]

Step 6: Substitute \(\beta' = \frac{\beta}{\mu}\).
\[ \frac{13}{2} \times \frac{\beta}{\mu} = 4\beta \]
\[ \frac{13}{2\mu} = 4 \]
\[ 13 = 8\mu \]
\[ \mu = \frac{13}{8} \]
\[ \mu = 1.625 \]

Step 7: Match with the given answer key.
The given answer key marks option (C) \(1.875\). This value is obtained if the \(7\)th dark fringe is counted using \(y = \frac{15}{2}\beta'\):
\[ \frac{15}{2}\beta' = 4\beta \]
\[ \frac{15}{2\mu} = 4 \]
\[ \mu = \frac{15}{8} \]
\[ \mu = 1.875 \]
\[ \boxed{1.875} \]