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

Concept:
In YDSE, position of the \(n^{\text{th}}\) bright fringe is: \[ y_n = y_0 + n\beta \] where \(y_0\) is the central maxima position and \(\beta\) is the fringe width. Fringe width is directly proportional to wavelength: \[ \beta \propto \lambda \] ip

Step 1: Find the fringe width for wavelength \(6000\text{\AA}\).
Given: \[ y_1=14.50\text{ mm}, \qquad y_{10}=16.75\text{ mm} \] Now, \[ y_{10}-y_1=9\beta \] So, \[ 16.75-14.50=9\beta \] \[ 2.25=9\beta \] \[ \beta=0.25\text{ mm} \] ip

Step 2: Find the central maxima position.
\[ y_1 = y_0+\beta \] So, \[ 14.50 = y_0+0.25 \] \[ y_0=14.25\text{ mm} \] ip

Step 3: Find new fringe width for \(5500\text{\AA}\).
Since \[ \beta'=\beta\cdot \frac{5500}{6000} \] \[ \beta'=0.25\cdot \frac{11}{12} \] \[ \beta' \approx 0.2292\text{ mm} \] ip

Step 4: Find new zero order and tenth order maxima positions.
Zero order maxima remains at: \[ y_0=14.25\text{ mm} \] Tenth order maxima: \[ y'_{10}=y_0+10\beta' \] \[ y'_{10}=14.25+10(0.2292) \] \[ y'_{10}=14.25+2.292\approx 16.55\text{ mm} \] ip Hence, the correct answer is:
\[ \boxed{(A)\ 14.25\text{ mm},\ 16.55\text{ mm}} \]