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MHT CET 2025 (20 Apr, Shift 2) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from MHT CET 2025 (20 Apr, Shift 2)

Q16MHT CET 2025 · 20 APR · SHIFT 2Medium

YDSE: Fringe Width & Fringe Positions

In Young's double slit experiment with monochromatic light of wavelength 600 nm, the distance between the slits is 10−3 m10^{-3}\text{ m}. For changing the fringe width by 3×10−5 m3 \times 10^{-5}\text{ m}

a. the screen is moved away from the slit by 5 cm.

b. the screen is moved 5 cm towards the slits.

c. the screen is moved 3 cm towards the slits.

d. the screen is moved away from the slits by 3 cm.

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Answer: (b)

  1. β=λDd⇒Δβ=λ ΔDd\beta = \frac{\lambda D}{d} \Rightarrow \Delta\beta = \frac{\lambda\,\Delta D}{d}
  2. ΔD=Δβ dλ=3×10−5×10−36×10−7=0.05 m=5 cm\Delta D = \frac{\Delta\beta\, d}{\lambda} = \frac{3\times10^{-5} \times 10^{-3}}{6\times10^{-7}} = 0.05\text{ m} = 5\text{ cm}
  3. Moving 5 cm away or 5 cm towards both change β\beta by this amount.
  4. So the answer is (b) both (a) and (b).
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