Question

A single slit diffraction pattern is formed with white light. For what wavelength of light the 4th secondary maximum in diffraction pattern coincides with the 3rd secondary maximum in the pattern of light of wavelength '$\lambda$'?

• A. $\frac{5\lambda}{7}$
• B. $\frac{7\lambda}{9}$
• C. $\frac{3\lambda}{4}$
• D. $\frac{9\lambda}{13}$

Hint:

Don't confuse Interference (YDSE) with Diffraction!
In YDSE, Maxima are at $n\lambda$.
In Single Slit, Maxima are at $(n + 0.5)\lambda$ and Minima are at $n\lambda$. This half-integer shift is crucial!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are dealing with a Fraunhofer single slit diffraction setup. We need to equate the angular positions of two specific secondary maxima generated by two different wavelengths.

Step 2: Detailed Explanation:
In a single slit diffraction pattern, the angular position ($\theta$) of the $n^{\text{th}}$ secondary maximum (bright band) is given by the formula:
$a \sin \theta = \left(n + \frac{1}{2}\right) \lambda$
where '$a$' is the width of the slit.
1. Position of the 4th secondary maximum of the unknown wavelength ($\lambda'$):
Here, $n = 4$.
$a \sin \theta_1 = \left(4 + \frac{1}{2}\right) \lambda' = 4.5 \lambda' = \frac{9}{2} \lambda'$
2. Position of the 3rd secondary maximum of the known wavelength ($\lambda$):
Here, $n = 3$.
$a \sin \theta_2 = \left(3 + \frac{1}{2}\right) \lambda = 3.5 \lambda = \frac{7}{2} \lambda$
3. Equating the positions:
The problem states that these two maxima physically coincide on the screen, which means their angular positions are identical ($\theta_1 = \theta_2$).
Therefore, we can equate the two expressions:
$\frac{9}{2} \lambda' = \frac{7}{2} \lambda$
The denominators (2) perfectly cancel out:
$9 \lambda' = 7 \lambda$
Solve for the unknown wavelength $\lambda'$:
$\lambda' = \frac{7\lambda}{9}$

Step 3: Final Answer:
The wavelength is $\frac{7\lambda}{9}$, matching option (b).