Question

Light of wavelength 500 nm falls on a single slit of width 0.1 mm. Angular position of 1st minimum is

• A. $\sin^{-1} (0.05)$
• B. $\sin^{-1} (0.2)$
• C. $\sin^{-1} (0.5)$
• D. $\sin^{-1} (0.0025)$
• E. $\sin^{-1} (0.005)$

Hint:

Always ensure all length units ($\lambda$ and $a$) are converted to the same base unit (like meters) before performing the calculation to avoid massive order-of-magnitude errors.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
When monochromatic light passes through a narrow single slit, it undergoes diffraction, creating a pattern of bright and dark fringes on a screen. The dark fringes (minima) occur due to destructive interference of wavelets originating from different parts of the slit.
Step 2: Key Formula or Approach:
The general condition for minima in a single slit diffraction pattern is given by:
\[ a \sin(\theta) = n\lambda \]
Where:
\(a\) = width of the slit
\(\theta\) = angular position of the minimum
\(n\) = order of the minimum (\(n = \pm 1, \pm 2, \dots\))
\(\lambda\) = wavelength of the light
Step 3: Detailed Explanation:
Given values:
Wavelength, \(\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}\)
Slit width, \(a = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 10^{-4} \text{ m}\)
Order of minimum, \(n = 1\) (for the first minimum)
We need to find the angular position \(\theta\). Using the formula:
\[ a \sin(\theta) = n\lambda \]
\[ \sin(\theta) = \frac{n\lambda}{a} \]
Substitute the values:
\[ \sin(\theta) = \frac{1 \times 500 \times 10^{-9}}{10^{-4}} \]
\[ \sin(\theta) = 500 \times 10^{-9 - (-4)} \]
\[ \sin(\theta) = 500 \times 10^{-5} \]
To convert this to a standard decimal:
\[ \sin(\theta) = 5.00 \times 10^2 \times 10^{-5} \]
\[ \sin(\theta) = 5 \times 10^{-3} \]
\[ \sin(\theta) = 0.005 \]
Therefore, the angle is:
\[ \theta = \sin^{-1}(0.005) \]
Step 4: Final Answer:
The angular position of the 1st minimum is \(\sin^{-1}(0.005)\).