Q143 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics

In young's double slit experiment with monochromatic light of wave length $6000 Å$. The fringe width is 3 mm . If the distance between the screen and slits is increased by. $50 \%$ and the distance between the slits 1 s decreased by $10 \%$, then the fringe width is

  • A
    12 mm
  • Correct answer
    B
    5 mm
  • C
    6 mm
  • D
    10 mm

Solution

In YDSE, $\beta_1=\frac{\lambda D}{d}=3 \mathrm{~mm}$ $\begin{aligned} & \beta_2=\frac{\lambda(1.5 \mathrm{D})}{(0.9 \mathrm{~d})}=\frac{5}{3} \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{5}{3} \beta_1 \\ & \therefore \quad \beta_2=\frac{5}{3} \times 3=5 \mathrm{~mm}\end{aligned}$

Hence, the correct answer is option: 'B'.

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