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Question

(A) The fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\lambda_2 = 600 \ nm > \lambda_1 = 400 \ nm$,it follows that $\beta_2 > \beta_1

Vedclass · Accepted Answer

$(A, B, C)$

Vedclass · Answer

(A) The fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\lambda_2 = 600 \ nm > \lambda_1 = 400 \ nm$,it follows that $\beta_2 > \beta_1$. Thus,$(A)$ is correct.The number of fringes $m$ in a distance $y$ is $m = \frac{y}{\beta}$. Since $\beta_2 > \beta_1$,we have $m_2 The position of the $n^{	ext{th}}$ maximum is $y_n = \frac{n \lambda D}{d}$. For the $3^{	ext{rd}}$ maximum of $\lambda_2$,$y = \frac{3 	imes 600 	imes D}{d} = \frac{1800 D}{d}$.The position of the $n^{	ext{th}}$ minimum is $y_n = \frac{(2n-1) \lambda D}{2d}$. For the $5^{	ext{th}}$ minimum of $\lambda_1$,$y = \frac{(2 	imes 5 - 1) 	imes 400 	imes D}{2d} = \frac{9 	imes 400 	imes D}{2d} = \frac{1800 D}{d}$.Since the positions are equal,they overlap. Thus,$(C)$ is correct.The angular separation is $	heta = \frac{\lambda}{d}$. Since $\lambda_1 Therefore,the correct options are $(A, B, C)$.

Similar Questions

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