Question

In a Young's double slit experiment wavelength of light used is $6000\text{\AA}$. The first order maxima and tenth order maxima fall at $14.50\text{ mm}$ and $16.75\text{ mm}$ from the particular reference point in the interference pattern respectively. If the wavelength is changed to $5500\text{\AA}$ then the position of zero order and tenth order maxima are respectively} [The other arrangements remaining same]

• A. $14.25\text{ mm}, 16.55\text{ mm}$
• B. $12.25\text{ mm}, 14.55\text{ mm}$
• C. $10.25\text{ mm}, 12.55\text{ mm}$
• D. $16.25\text{ mm}, 18.55\text{ mm}$

Hint:

In YDSE: \[ y_n=y_0+n\beta,\qquad \beta \propto \lambda \] So when wavelength changes, central maxima stays fixed but fringe width changes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The position of the $n^{th}$ order maxima is $y_n = y_0 + n\beta$, where $y_0$ is the position of the zero-order maxima and $\beta = \frac{\lambda D}{d}$ is the fringe width.
Step 2: Key Formula or Approach:
Fringe width $\beta = \frac{y_{n2} - y_{n1}}{n_2 - n_1}$.
Ratio of fringe widths: $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.
Step 3: Detailed Explanation:
Initially, for $\lambda_1 = 6000\text{\AA}$:
\[ \beta_1 = \frac{16.75 - 14.50}{10 - 1} = \frac{2.25}{9} = 0.25\text{ mm} \]
Position of first maxima: $14.50 = y_0 + 1(0.25) \implies y_0 = 14.25\text{ mm}$.
The position of zero-order maxima $y_0$ remains $14.25\text{ mm}$ regardless of wavelength.
New fringe width for $\lambda_2 = 5500\text{\AA}$:
\[ \beta_2 = \beta_1 \times \frac{5500}{6000} = 0.25 \times \frac{11}{12} \approx 0.22916\text{ mm} \]
New position of tenth maxima:
\[ y_{10}' = 14.25 + 10(0.22916) = 14.25 + 2.29 = 16.54\text{ mm} \approx 16.55\text{ mm} \]
Step 4: Final Answer:
The positions are $14.25\text{ mm}$ and $16.55\text{ mm}$.