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MHT CET · Physics · Wave Optics

In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is at 1.5 m away from the lens. If wavelength of light used is \(4500 Å\), then the distance between the first minimum on either side of the central maximum is [ \(\theta\) is small and measured in radian.]

  1. A 1.5 mm
  2. B 2.25 mm
  3. C 3.25 mm
  4. D 4.5 mm
Verified Solution

Answer & Solution

Correct Answer

(D) 4.5 mm

Step-by-step Solution

Detailed explanation

\( \text{Distance} = \frac{2D\lambda}{a} \) \( = \frac{2 \times 1.5 \text{ m} \times 4500 \times 10^{-10} \text{ m}}{0.3 \times 10^{-3} \text{ m}} \)