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In a Young's double slit experiment, light of wavelength \(5900 Å\) is used. When the slits are \(2 \mathrm{~mm}\) apart, the fringe width is \(1.2 \mathrm{~mm}\). If the slit separation is increased to one and half times the previous value, then the fringe width will be
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The correct answer is:
\(0.8 \mathrm{~mm}\)
In a Young's double slit experiment, wavelength of light, \(\lambda=5900 Å\) and distance between the slits, \(d_1=d=2 \times 10^{-3} \mathrm{~m}\),
Fringe width, \(\beta_1=1.2 \times 10^{-3} \mathrm{~m}\) and \(d_2=\frac{3}{2} \mathrm{~d}\)
As, the fringe width,
\(\beta=\frac{\lambda D}{d}\)
So, \(\quad \beta_1=\frac{\lambda_1 D_1}{d_1}\) and \(\beta_2=\frac{\lambda_2 D_2}{d_2}\)
Given, \(\lambda_1=\lambda_2\) and \(D_1=D_2 \Rightarrow \frac{\beta_1}{\beta_2}=\frac{d_2}{d_1}\)
Putting the given values, we get
\(\Rightarrow \frac{1.2 \times 10^{-3}}{\beta_2}=\frac{\frac{3}{2} d}{d} \Rightarrow \beta_2=0.8 \mathrm{~mm}\)
Hence, the correct option is (b).
Fringe width, \(\beta_1=1.2 \times 10^{-3} \mathrm{~m}\) and \(d_2=\frac{3}{2} \mathrm{~d}\)
As, the fringe width,
\(\beta=\frac{\lambda D}{d}\)
So, \(\quad \beta_1=\frac{\lambda_1 D_1}{d_1}\) and \(\beta_2=\frac{\lambda_2 D_2}{d_2}\)
Given, \(\lambda_1=\lambda_2\) and \(D_1=D_2 \Rightarrow \frac{\beta_1}{\beta_2}=\frac{d_2}{d_1}\)
Putting the given values, we get
\(\Rightarrow \frac{1.2 \times 10^{-3}}{\beta_2}=\frac{\frac{3}{2} d}{d} \Rightarrow \beta_2=0.8 \mathrm{~mm}\)
Hence, the correct option is (b).
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