Wave Optics

JEE Main level, single correct, medium

16 of 191

Question 16 of 191

The interference pattern is obtained with two coherent light sources of intensity ratio 4:14:1. And the ratio Imax⁡+Imin⁡Imax⁡−Imin⁡\dfrac{{I}_{\max }+{I}_{\min }}{{I}_{\max }-{I}_{\min }} is 5x\dfrac{5}{x}. Then, the value of xx will be equal to :

Show answer and solution

Answer: (B) 44

The ratio of the intensities is I1I2=41\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{4}{1}
After taking square root of above equation on both sides, we get, I1I2=21\dfrac{\sqrt{{I}_{1}}}{\sqrt{{I}_{2}}}=\dfrac{2}{1}
Apply componendo and dividendo,
⇒I1+I2I1−I2=2+12−1=31\Rightarrow \dfrac{\sqrt{{I}_{1}}+\sqrt{{I}_{2}}}{\sqrt{{I}_{1}}-\sqrt{{I}_{2}}}=\dfrac{2+1}{2-1}=\dfrac{3}{1}
⇒(I1+I2I1−I2)2=Imax⁡Imin⁡=91⇒Imax+IminImax−Imin=108=54\Rightarrow {\left(\dfrac{\sqrt{{I}_{1}}+\sqrt{{I}_{2}}}{\sqrt{{I}_{1}}-\sqrt{{I}_{2}}}\right)}^{2}=\dfrac{{I}_{\max }}{{I}_{\min }}=\dfrac{9}{1} \Rightarrow \dfrac{{I}_{max}+{I}_{min}}{{I}_{max}-{I}_{min}}=\dfrac{10}{8}=\dfrac{5}{4}
Therefore, the answer is 44.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Wave optics: YDSE, diffraction and polarisation simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Wave Optics (Interference of Light), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.