Question:medium

In Young’s double slit experiment, the ratio of intensities of maxima and minima is \(25:9\). The ratio of intensities of two slits is:

Show Hint

Remember: \[ \frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1}+\sqrt{I_2}} {\sqrt{I_1}-\sqrt{I_2}} \right)^2 \] Take square root first to simplify calculations quickly.
Updated On: Jun 17, 2026
  • \(18:3\)
  • \(4:1\)
  • \(8:1\)
  • \(16:1\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Recall the max and min intensity formulas.
In a double slit pattern the brightest and darkest spots are \[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2,\quad I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \]

Step 2: Use the given ratio.
We are told $\dfrac{I_{max}}{I_{min}} = \dfrac{25}{9}$. Take the square root of both sides. \[ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = \frac{5}{3} \]
Step 3: Use simple letters.
Let $a = \sqrt{I_1}$ and $b = \sqrt{I_2}$. \[ \frac{a + b}{a - b} = \frac{5}{3} \]
Step 4: Cross multiply.
\[ 3(a + b) = 5(a - b) \] \[ 3a + 3b = 5a - 5b \]
Step 5: Solve for the ratio of a to b.
\[ 8b = 2a \quad\Rightarrow\quad a = 4b \]
Step 6: Get the intensity ratio.
Square the relation since $I = (\text{root})^2$. \[ \frac{I_1}{I_2} = \frac{a^2}{b^2} = 16:1 \] \[ \boxed{16:1} \]
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