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.]
- A 1.5 mm
- B 2.25 mm
- C 3.25 mm
- D 4.5 mm
Answer & Solution
Correct Answer
(D) 4.5 mm
Step-by-step Solution
Detailed explanation
\( \text{Distance} = \frac{2D\lambda}{a} \) \( = \frac{2 \times 1.5 \text{ m} \times 4500 \times 10^{-10} \text{ m}}{0.3 \times 10^{-3} \text{ m}} \)
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