COMEDK · Physics · 26. Wave Optics
The wavelength of a monochromatic light which is used in single slit diffraction is 800 nm . The width of the single slit for which the first minimum appears at \(\theta=45^{\circ}\) on the screen will be:
- A \(1.3 \mu \mathrm{~m}\)
- B \(1.13 \mu \mathrm{~m}\)
- C \(2.13 \mu \mathrm{~m}\)
- D \(1.23 \mu \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(1.13 \mu \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The condition for the first minimum in single slit diffraction is given by the formula \(a \sin \theta = n \lambda\), where \(a\) is the width of the slit, \(\theta\) is the angle of diffraction, \(n\) is the order of the minimum, and \(\lambda\) is the wavelength of the light.…
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