Physics Circle home Student Login

Home›Chapters›Wave Optics PYQs›YDSE: Fringe Width & Fringe Positions›Question 76

NEET 1992 (AIPMT) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from NEET 1992 (AIPMT)

Q76NEET 1992 · AIPMTEasy

YDSE: Fringe Width & Fringe Positions

In Young's double slit experiment carried out with light of wavelength (λ)=5000(\lambda) = 5000 Å, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x=0x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at xx equal to

Show answer and solution

Answer: (b)

  1. Fringe width β=λDd=5×10−7×22×10−4=5×10−3 m=0.5 cm\beta = \frac{\lambda D}{d} = \frac{5\times10^{-7} \times 2}{2\times10^{-4}} = 5\times10^{-3}\text{ m} = 0.5\text{ cm}
  2. Third maximum: x=3β=1.5 cmx = 3\beta = 1.5\text{ cm}
  3. So the answer is (b) 1.5 cm1.5\text{ cm}.
Report a mistake

More questions on YDSE: Fringe Width & Fringe Positions