Question

In a single slit diffraction experiment, if the third minimum for light of wavelength 4800 Å coincides with the second secondary maximum of another light of wavelength $\lambda$, then the value of is

• A. 6000 Å
• B. 5500 Å
• C. 5760 Å
• D. 4320 Å \bigskip

Hint:

Diffraction maxima are approximately located between successive minima.

Answer 1

The Correct Option is C

Concept: For single slit diffraction: \[ \text{minima: } a\sin\theta = n\lambda \] Secondary maxima approximately lie between minima. ---

Step 1: Third minimum condition.
\[ n=3 \Rightarrow a\sin\theta = 3 \times 4800 \] \[ = 14400 \] ---

Step 2: Second secondary maximum approximation.
Second secondary maximum occurs near: \[ a\sin\theta \approx \frac{5}{2}\lambda \] ---

Step 3: Equate positions.
\[ 3 \times 4800 = \frac{5}{2}\lambda \] \[ 14400 = \frac{5}{2}\lambda \Rightarrow \lambda = 5760~Å \] --- Final Answer: \[ (C) \]