AP EAMCET · PHYSICS · Wave Optics
In Young's double slit experiment, if the distance between the slits is 2 mm and the distance of the screen from the slits is 100 cm, the fringe width is 0.36 mm. If the distance between the slits is decreased by 0.5 mm and the distance of the screen from the slits is increased by 50 cm, the fringe width becomes
- A 0.84 mm
- B 0.96 mm
- C 0.48 mm
- D 0.72 mm
Answer & Solution
Correct Answer
(D) 0.72 mm
Step-by-step Solution
Detailed explanation
\(d_1 = 2 \text{ mm}, D_1 = 100 \text{ cm}, \beta_1 = 0.36 \text{ mm}\) \(d_2 = 2 - 0.5 = 1.5 \text{ mm}\) \(D_2 = 100 + 50 = 150 \text{ cm}\) \(\beta_2 = \beta_1 \frac{D_2}{D_1} \frac{d_1}{d_2}\)…
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