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Question

(D) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.Given: $\lambda = 5896 	imes 10^{-10} \, m$,$D = 1 \, m$,$d = 0.2

Vedclass · Accepted Answer

$0.295$

Vedclass · Answer

(D) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.Given: $\lambda = 5896 	imes 10^{-10} \, m$,$D = 1 \, m$,$d = 0.2 \, cm = 2 	imes 10^{-3} \, m$.Calculating $\beta$ in air:$\beta = \frac{5896 	imes 10^{-10} 	imes 1}{2 	imes 10^{-3}} = 2948 	imes 10^{-7} \, m = 0.2948 \, mm \approx 0.295 \, mm$.When the system is immersed in water,the wavelength changes to $\lambda_w = \frac{\lambda}{\mu_w}$.Therefore,the new fringe width $\beta_w = \frac{\beta}{\mu_w} = \frac{0.2948}{4/3} = 0.2948 	imes 0.75 = 0.2211 \, mm$.Since the question asks for the fringe width in air as the primary value,the closest option is $0.295 \, mm$.

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