Physics Q43 – JEE-MAIN 2025

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:

Clue 1 of 3: Intensity and Slit Width

Recall that the intensity of light from a slit is proportional to its width, and the intensity in an interference pattern is proportional to the square of the amplitude.

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Ninja Strategy•Estimate the Ratio

Estimate the value of $\sqrt{2} \approx 1.414$ and calculate the approximate numerical value of the ratio for each option. The correct answer, $(3 + 2 \sqrt{2}) : (3 - 2 \sqrt{2})$ , approximates to $(3 + 2.828) : (3 - 2.828) = 5.828 : 0.172 \approx 33.88 : 1$ . This large ratio helps eliminate options with small numerical values.

🧠 Full Solution Path

Step 1: Relate Slit Width to Intensity✦ Active

The intensity of light from a slit is proportional to its width. Let the widths of the two slits be $w_{1}$ and $w_{2}$ . Given that one slit width is half of the other, we can set $w_{1} = w$ and $w_{2} = 2 w$ . Consequently, the intensities of light from the two slits will be $I_{1} = I_{0}$ and $I_{2} = 2 I_{0}$ (or vice versa, which yields the same final ratio due to symmetry).

Step 2: Apply Interference Intensity Formulas○ Expand

The maximum and minimum intensities in an interference pattern are given by the formulas:

I_{m a x} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}

I_{m i n} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}

The ratio of maximum to minimum intensity is therefore:

\frac{I_{m a x}}{I_{m i n}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}

💡 Teacher's Secret Hint

Remember that the amplitude is proportional to the square root of intensity, $A \propto \sqrt{I}$ .

Step 3: Substitute Values and Simplify○ Expand

Substitute $I_{1} = I_{0}$ and $I_{2} = 2 I_{0}$ into the ratio formula:

\frac{I_{m a x}}{I_{m i n}} = {(\frac{\sqrt{I_{0}} + \sqrt{2 I_{0}}}{\sqrt{I_{0}} - \sqrt{2 I_{0}}})}^{2}

Factor out $\sqrt{I_{0}}$ from the numerator and denominator:

\frac{I_{m a x}}{I_{m i n}} = {(\frac{\sqrt{I_{0}} (1 + \sqrt{2})}{\sqrt{I_{0}} (1 - \sqrt{2})})}^{2} = {(\frac{1 + \sqrt{2}}{1 - \sqrt{2}})}^{2}

Now, expand the squares in the numerator and denominator:

\frac{I_{m a x}}{I_{m i n}} = \frac{(1 + \sqrt{2})^{2}}{(1 - \sqrt{2})^{2}} = \frac{1^{2} + (\sqrt{2})^{2} + 2 (1) (\sqrt{2})}{1^{2} + (\sqrt{2})^{2} - 2 (1) (\sqrt{2})} = \frac{1 + 2 + 2 \sqrt{2}}{1 + 2 - 2 \sqrt{2}} = \frac{3 + 2 \sqrt{2}}{3 - 2 \sqrt{2}}

Thus, the ratio of maximum to minimum intensity is $(3 + 2 \sqrt{2}) : (3 - 2 \sqrt{2})$ . This corresponds to option 2.

💡 Teacher's Secret Hint

Be careful with the algebraic expansion and simplification of terms involving square roots.

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Physics Question 43 – JEE-MAIN 2025

Clue 1 of 3: Intensity and Slit Width

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