Question:

A slit of width ‘a’ is illuminated by red light of wavelength 6500 A°. If the first diffraction minimum falls at 30°, then the value of ‘a’ is ________

Updated On: Apr 10, 2025
  • 6.5 × 10-4 mm
  • 1.3 micron
  • 3250 A°
  • 2.6 × 10-4 cm
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The Correct Option is B

Approach Solution - 1

Using the diffraction formula for the first minimum:

\[ a \sin \theta = n \lambda \]

Given:

\[ \theta = 30°, \quad n = 1, \quad \lambda = 6500 \, \text{Å} = 6.5 \times 10^{-7} \, \text{m} \]

Substitute the values:

\[ a \sin 30° = 6.5 \times 10^{-7} \, \text{m} \]

\[ a = \frac{6.5 \times 10^{-7} \, \text{m}}{0.5} = 1.3 \times 10^{-6} \, \text{m} \]

Convert to microns:

\[ a = 1.3 \, \text{micron} \]

Thus, the correct answer is Option (B).

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Approach Solution -2

Step 1: Identify Formula

Condition for first diffraction minimum: $a \sin\theta = \lambda$

Step 2: List Given Values

$\lambda = 6500 \ A°$
$\theta = 30°$

Step 3: Substitute Values

$a \sin(30°) = 6500 \ A°$

Step 4: Solve for 'a'

$a = \frac{6500 \ A°}{\sin(30°)} = \frac{6500 \ A°}{1/2} = 13000 \ A°$

Step 5: Convert to meters

$a = 13000 \times 10^{-10} \ m = 1.3 \times 10^{-6} \ m$

Step 6: Convert to microns

$a = 1.3 \times 10^{-6} \ m = 1.3 \ micron$

Final Answer: The final answer is $1.3 \ micron$

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