Question

A source S is placed at distance 3λ from end E₃ of a plane mirror. Screen E₁E₂ is placed as shown in figure where detector can observe the interference pattern including point E₂. E₃, S and E₂ are collinear and line joining E₃, S and E₂ is perpendicular to mirror. Number of maximas observed by the detector is

Answer 1

7

Answer 2

The setup involves interference between waves from a source S and its image S' formed by the plane mirror. Let the mirror be along the y-axis and the line E₃SE₂ along the x-axis. Let E₃ be at the origin (0,0). Then S is at (3λ, 0). The image S' is at (-3λ, 0). The screen is the line segment E₁E₂. Point E₂ is on the x-axis at some position L, where L > 3λ. Point E₁ is above E₃.

At point E₂, which is on the x-axis at x=L, the path difference $\Delta r$ is the difference between the distance from S' to E₂ and the distance from S to E₂. $PS' = \sqrt{(L - (-3\lambda))^2 + 0^2} = \sqrt{(L+3\lambda)^2} = L+3\lambda$ (since L > 3λ) $PS = \sqrt{(L - 3\lambda)^2 + 0^2} = L-3\lambda$ (since L > 3λ) $\Delta r_{E_2} = PS' - PS = (L+3\lambda) - (L-3\lambda) = 6\lambda$. For maxima, $\Delta r = n\lambda$. So, at E₂, $n\lambda = 6\lambda$, which means $n=6$.

At point E₁, which is directly above E₃ (let's assume its coordinates are (0, h) for some height h), the path difference is: $PS' = \sqrt{(0 - (-3\lambda))^2 + h^2} = \sqrt{(3\lambda)^2 + h^2}$ $PS = \sqrt{(0 - 3\lambda)^2 + h^2} = \sqrt{(-3\lambda)^2 + h^2} = \sqrt{(3\lambda)^2 + h^2}$ $\Delta r_{E_1} = PS' - PS = 0$. For maxima, $\Delta r = n\lambda$. So, at E₁, $n\lambda = 0$, which means $n=0$.

The detector observes the interference pattern along the screen E₁E₂. The path difference varies from 0 at E₁ to 6λ at E₂. The number of maxima corresponds to the integer values of n in the range [0, 6]. These values are 0, 1, 2, 3, 4, 5, 6. Therefore, there are a total of 7 maxima observed.