TS EAMCET · Physics · Wave Optics
In a double slit interference experiment, the fringe width obtained with a light of wavelength \(5900 Å\) was \(1.2 \mathrm{~mm}\) for parallel narrow slits placed \(2 \mathrm{~mm}\) apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is
- A \(0.9 \mathrm{~mm}\)
- B \(0.8 \mathrm{~mm}\)
- C \(1.8 \mathrm{~mm}\)
- D \(1.6 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(0.8 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
By Young's double slit interference experiment \[ \beta=\frac{\lambda D}{d} \] The given \[ \begin{aligned} & \beta_1=1.2 \mathrm{~mm} \\ & \frac{d_2}{d_1}=1 \frac{1}{2}=1.5 \end{aligned} \] So, \(\quad \beta \propto \frac{1}{d}\)…
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