JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 10)
In a Young's double slits experiment, the ratio of amplitude of light coming from slits is $$2: 1$$. The ratio of the maximum to minimum intensity in the interference pattern is:
25 : 9
9 : 1
9 : 4
2 : 1
Explanation
Given the amplitude ratio, $$\frac{A_1}{A_2} = \frac{2}{1}$$, we can find the maximum and minimum intensities using the formula:
$$\frac{I_{max}}{I_{min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$$
Substituting the given amplitude ratio:
$$\frac{I_{max}}{I_{min}} = \left(\frac{2 + 1}{2 - 1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$$
Thus, the ratio of maximum to minimum intensity in the interference pattern is:
$$\frac{I_{max}}{I_{min}} = 9:1$$
$$\frac{I_{max}}{I_{min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$$
Substituting the given amplitude ratio:
$$\frac{I_{max}}{I_{min}} = \left(\frac{2 + 1}{2 - 1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$$
Thus, the ratio of maximum to minimum intensity in the interference pattern is:
$$\frac{I_{max}}{I_{min}} = 9:1$$
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