Question

A single slit diffraction pattern is formed with light of wavelength \(6384 \mathop {\rm{A}}\limits^{\rm{o}}\). The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' \(\lambda_0\) '. The value of ' \(\lambda_0\) ' is

• A. \(4242\overset{\mathrm{o}}{\mathrm{A}}\)
• B. \(4560\overset{\mathrm{o}}{\mathrm{A}}\)
• C. \(5474\overset{\mathrm{o}}{\mathrm{A}}\)
• D. \(6384\overset{\mathrm{o}}{\mathrm{A}}\)

Hint:

Secondary maxima in a single slit occur at a sin(theta) = (n + 1/2) lambda; equate the angles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In single slit diffraction, the \(n\)-th secondary maximum satisfies \(a\sin\theta=\left(n+\dfrac12\right)\lambda\).

Step 2: Write the condition
Second secondary maximum for 6384 \(\text{\AA}\): \(a\sin\theta=\dfrac52\times6384\).
Third secondary maximum for \(\lambda_0\): \(a\sin\theta=\dfrac72\lambda_0\).

Step 3: Equate
\(\dfrac72\lambda_0=\dfrac52\times6384\)
\(\lambda_0=\dfrac{5\times6384}{7}=5\times912=4560\ \text{\AA}\).

Final Answer:
The wavelength is 4560 angstrom. \[ \boxed{\text{(B) }4560\,\text{\AA}} \]