Question

in young's double slit expmnt. the ratio of max. and min. intensities of fringes are 4:1.what are the amplitudes of the coherent sources?

please answer................

Answer 1

Answer on Question Question #43089, Physics, Optics

Question: "in young's double slit expmnt. the ratio of max. and min. intensities of fringes are 4:1.what are the amplitudes of the coherent sources?".

Solution:

The total instantaneous electric field $E$ at the point $P$ on the screen is equal to the vector sum of the two sources: $\vec{E} = \vec{E}_1 + \vec{E}_2$. Maximum intensity is $(E_1 + E_2)$ and minimum intensity is $(E_1 - E_2)$.

The intensity $I$ of the light at $P$ is $\approx E^2$.

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{(E_1 + E_2)^2}{(E_1 - E_2)^2} = \frac{4}{1}$$ $$\frac{E_1 + E_2}{E_1 - E_2} = 2$$

Thus we have $E_1 = 3E_2$

Answer: $E_1 = 3E_2$

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