Question

In a single-slit diffraction experiment, the slit is illuminated by light of two wavelengths \(\lambda_1\) and \(\lambda_2\). It is observed that the 2nd order diffraction minimum for \(\lambda_1\) coincides with the 3rd diffraction minimum for \(\lambda_2\). Then:

• A. \(\frac{\lambda_1}{\lambda_2} = \frac{2}{3}\)
• B. \(\frac{\lambda_1}{\lambda_2} = \frac{5}{7}\)
• C. \(\frac{\lambda_1}{\lambda_2} = \frac{3}{2}\)
• D. \(\frac{\lambda_1}{\lambda_2} = \frac{7}{5}\)

Hint:

In diffraction problems, use the diffraction condition equation and compare the angles for different orders to relate the wavelengths involved.

Answer 1

The Correct Option is C

Step 1: In single-slit diffraction, the condition for the \( m \)-th diffraction minimum is given by:

\[ a \sin \theta_m = m\lambda, \]

where \( a \) is the width of the slit, \( \lambda \) is the wavelength of light, and \( m \) is the diffraction order.

Step 2: For the 2nd order minimum of \( \lambda_1 \) and the 3rd order minimum of \( \lambda_2 \), we can set the angle for both minima equal since they coincide:

\[ 2\lambda_1 = 3\lambda_2 \]

Thus:

\[ \frac{\lambda_1}{\lambda_2} = \frac{3}{2}. \]