Question

In Young's double slit experiment, using monochromatic light of wave length $\mathsf{\lambda}$ , the intensity of light at a point on the screen where path difference is $\mathsf{\lambda}$ is k units. What is the intensity of light at a point where path difference is $\mathsf{ \frac{\lambda}{3} \\ }$ ?

Answer 1

$\sf{Path \: difference \: \: x_{1}} \: = \lambda$

$\sf{Intensity \: I_{1} = K}$

$\sf{Path \: difference \: \: x_{2}} \: = \frac{ \lambda}{3}$

$\sf{Intensity \: I_{2} = \: ? }$

$\\ \sf{Phase \: difference \: \phi _{1} \: = \frac{2 \pi}{ \lambda} x_1 \: = \frac{2 \pi}{ \lambda} ( \lambda) = 2 \pi } \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \phi _{2} \: = \frac{2 \pi}{ \lambda} \bigg( \frac{ \lambda}{3} \bigg) \: = \frac{2 \pi}{ 3} }$

$\rm{We \: know \: that, \boxed{ \boxed{I = 4 \: I_{0} \: {cos}^{2} \bigg( \frac{\phi}{2} \bigg)}}}$

$\frac{I _1}{I _2} = \frac{ \: {cos}^{2} \bigg( \frac{\phi_1}{2} \bigg)}{\: {cos}^{2} \bigg( \frac{\phi_2}{2} \bigg)}$

$\implies \frac{K}{I _2} = \frac{ \: {cos}^{2} \bigg( \frac{2\pi}{2} \bigg)}{\: {cos}^{2} \bigg( \frac{2 \pi}{3(2)} \bigg)}$

$\implies \frac{K}{I _2} = \frac{ 1}{ \frac{1}{4} }$

$\implies \frac{K}{I _2} = 4$

$\implies \frac{K}{4} = I _2$

Answer 2

Answer:

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Explanation:

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