MHT CET · Physics · Wave Optics
In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(x\) units, \(\lambda\). being the wavelength of light used. The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be \(\left(\cos 2 \pi=1, \cos \frac{\pi}{2}=0\right)\)
- A \(\frac{\mathrm{x}}{4}\)
- B \(\frac{x}{2}\)
- C x
- D zero
Answer & Solution
Correct Answer
(B) \(\frac{x}{2}\)
Step-by-step Solution
Detailed explanation
When path difference \(=\lambda\)
Phase difference
\(\begin{aligned}
& \quad \Delta \phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi \\
& \\
& I=4 I_0 \cos ^2\left(\frac{\phi}{2}\right) \\
& \\
& x=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)=4 I_0 \\
& \therefore \quad \\
& x=4 I_0
\end{aligned}\)
When path difference \(=\frac{\lambda}{4}\)
Phase difference
\(\begin{aligned}
& \Delta \delta=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2} \\
& I=4 I_0 \cos ^2\left(\frac{\frac{\pi}{2}}{2}\right)=2 I_0
\end{aligned}\)
Intensity, \(x^{\prime}=\frac{x}{2}\)
Phase difference
\(\begin{aligned}
& \quad \Delta \phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi \\
& \\
& I=4 I_0 \cos ^2\left(\frac{\phi}{2}\right) \\
& \\
& x=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)=4 I_0 \\
& \therefore \quad \\
& x=4 I_0
\end{aligned}\)
When path difference \(=\frac{\lambda}{4}\)
Phase difference
\(\begin{aligned}
& \Delta \delta=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2} \\
& I=4 I_0 \cos ^2\left(\frac{\frac{\pi}{2}}{2}\right)=2 I_0
\end{aligned}\)
Intensity, \(x^{\prime}=\frac{x}{2}\)
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