Question

In Young’s double slit experiment using a monochromatic light of wavelength  , the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

Answer 1

Explanation:

I' = I cos²(Ф/2)

I/2 = I cos²(Ф/2)

cos(Ф/2) =  ± 1/√2

The value is obtained for  π/4 , 3π/4 , (2n + 1)π/4

Ф/2 = (2n + 1)π/4

Ф = (2n + 1)π/2

path difference = (2n + 1)π λ/4

Answer 2

Answer:

Answer:Corresponding to any point having half the peak intensity is $\frac{\lambda}{4}(2 n+1)$.

Explanation:

Step 1: Having or taking part in the same relationship (such as kind, degree, position, correspondence, or function), particularly in relation to similar or identical wholes (such as geometric figures or sets), such as the corresponding triangle parts.The degree, volume, or magnitude of something is its intensity. Examples include fire, emotion, weather, labour, or passion. The term "intensity" is occasionally linked to fervour, fire, and violence. It is employed when describing the intensity of things like a flame or perhaps a love engagement.

Step 2: The rate at which light disperses over a surface of a specific region some distance from a source is referred to as intensity.

The intensity changes depending on the source's force and distance from it.The intensity of a wave is the amount of energy it transports over a surface in a measure of time and area. It is also equal to the energy density times the wave speed. Watts per square metre are typically used to quantify it.

A wave's power and amplitude will affect intensity.

$\begin{aligned}& \frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\mathrm{m}} \cos ^2\left(\frac{\phi}{2}\right) \\& \Rightarrow \cos \left(\frac{\phi}{2}\right)=\frac{1}{\sqrt{2}} \\& \Rightarrow \frac{\phi}{2}=\frac{\pi}{4} \\& \Rightarrow \phi=\frac{\pi}{2}(2 \mathrm{n}+1) \\& \Rightarrow \Delta \mathrm{x}=\frac{\lambda}{2 \pi} \phi=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}(2 \mathrm{n}+1)=\frac{\lambda}{4}(2 \mathrm{n}+1)\end{aligned}$

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