Question

In Young's double slit experiment, the \(10^{th}\) maximum of wavelength \(\lambda_1\) is at a distance \(y_1\) from the central maximum. When the wavelength of the source is changed to \(\lambda_2\), the \(5^{th}\) maximum is at a distance of \(y_2\) from the central maximum. The ratio \(\left(\dfrac{y_1}{y_2}\right)\) is

• A. \(\dfrac{2\lambda_1}{\lambda_2}\)
• B. \(\dfrac{2\lambda_2}{\lambda_1}\)
• C. \(\dfrac{\lambda_1}{2\lambda_2}\)
• D. \(\dfrac{\lambda_2}{2\lambda_1}\)

Hint:

Position of \(n^{th}\) bright fringe: \(y_n = n\frac{\lambda D}{d}\). Ratio depends only on \(n\) and \(\lambda\).

Answer 1

The Correct Option is A

Step 1: Use position of maxima formula.
In YDSE, \(n^{th}\) bright fringe position is:
\[ y_n = n\frac{\lambda D}{d} \]
Step 2: Apply for first case.
For \(10^{th}\) maximum with wavelength \(\lambda_1\):
\[ y_1 = 10\frac{\lambda_1 D}{d} \]
Step 3: Apply for second case.
For \(5^{th}\) maximum with wavelength \(\lambda_2\):
\[ y_2 = 5\frac{\lambda_2 D}{d} \]
Step 4: Take ratio.
\[ \frac{y_1}{y_2} = \frac{10\lambda_1 D/d}{5\lambda_2 D/d} = \frac{10\lambda_1}{5\lambda_2} = \frac{2\lambda_1}{\lambda_2} \]
Final Answer: \[ \boxed{\dfrac{2\lambda_1}{\lambda_2}} \]