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NEET
Exam
NEET
Subject
Physics
Chapter
Wave Optics

NEET · Physics · Wave Optics

In a Young's double slit experiment, the initial angular fringe...

In a Young's double slit experiment, the initial angular fringe width is 0.1∘0.1^{\circ} and the corresponding linear fringe width is 2 mm2 \text{ mm}. The experimental setup is modified such that the new angular fringe width becomes 0.15∘0.15^{\circ} and the new linear fringe width becomes 4 mm4 \text{ mm}. The ratio of the new screen distance to the initial screen distance (D2D1)\left(\dfrac{D_2}{D_1}\right) is:

Options

  1. A33
  2. B13\dfrac{1}{3}
  3. C43\dfrac{4}{3}
  4. D22

Correct answer

C. 43\dfrac{4}{3}

Step-by-step solution

The linear fringe width β\beta and the angular fringe width θ\theta are related by the equation:

β=θD\beta = \theta D

where DD is the distance of the screen from the slits.

Rearranging for DD, we get:

D=βθD = \dfrac{\beta}{\theta}

The ratio of the new screen distance to the initial screen distance is:

D2D1=(β2β1)×(θ1θ2)\dfrac{D_2}{D_1} = \left(\dfrac{\beta_2}{\beta_1}\right) \times \left(\dfrac{\theta_1}{\theta_2}\right)

Substituting the given values:
β1=2 mm\beta_1 = 2 \text{ mm}, β2=4 mm\beta_2 = 4 \text{ mm}
θ1=0.1∘\theta_1 = 0.1^{\circ}, θ2=0.15∘\theta_2 = 0.15^{\circ}

D2D1=(42)×(0.10.15)=2×(23)=43\dfrac{D_2}{D_1} = \left(\dfrac{4}{2}\right) \times \left(\dfrac{0.1}{0.15}\right) = 2 \times \left(\dfrac{2}{3}\right) = \dfrac{4}{3}

Answer: 43\dfrac{4}{3}
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