Wave Optics — Easy JEE physics MCQ
In YDSE, the slit separation is $1$ mm, screen distance is $1$ m, and wavelength is $600$ nm. The fringe width is:
- A. $0.6$ mm
- B. $0.3$ mm
- C. $1.2$ mm
- D. $6$ mm
Solution
$\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{10^{-3}} = 6 \times 10^{-4}$ m $= 0.6$ mm