Question

Light of wavelength 5000 \r{A is incident normally on a slit of width $2.5 \times 10^{-4}$ cm. The angular position of second minimum from the central maximum is}

• A. $\sin^{-1} \left( \frac{1}{5} \right)$
• B. $\sin^{-1} \left( \frac{2}{5} \right)$
• C. $\left( \frac{\pi}{3} \right)$
• D. $\left( \frac{\pi}{6} \right)$
• E. $\left( \frac{\pi}{4} \right)$

Hint:

Be extremely careful with powers of 10! Always convert everything to cm or meters before starting. A common error is mixing Angstroms ($10^{-10}$ m) with centimeters ($10^{-2}$ m) without proper adjustment.

Answer 1

The Correct Option is B

Concept: This problem relates to the diffraction of light through a single slit.
• Diffraction Condition for Minima: For a slit of width $a$, the angular positions ($\theta$) of the minima are given by: \[ a \sin \theta = n\lambda \] Where $n$ is the order of the minimum ($n = 1, 2, 3, \dots$).
• Central Maximum: The region between the first minima on either side.

Step 1: Convert all units to a standard form.

• Wavelength $\lambda = 5000 \text{ \r{A}} = 5000 \times 10^{-8} \text{ cm} = 5 \times 10^{-5} \text{ cm}$.
• Slit width $a = 2.5 \times 10^{-4} \text{ cm}$.
• Order of minimum $n = 2$ (second minimum).

Step 2: Apply the diffraction formula.
Using $a \sin \theta = n\lambda$: \[ (2.5 \times 10^{-4}) \sin \theta = 2 \times (5 \times 10^{-5}) \] \[ (2.5 \times 10^{-4}) \sin \theta = 10 \times 10^{-5} = 10^{-4} \]

Step 3: Solve for the angular position $\theta$.
\[ \sin \theta = \frac{10^{-4}}{2.5 \times 10^{-4}} = \frac{1}{2.5} \] To express as a simple fraction: \[ \sin \theta = \frac{1}{2.5} = \frac{10}{25} = \frac{2}{5} \] \[ \theta = \sin^{-1} \left( \frac{2}{5} \right) \]