KCET · Physics · Wave Optics
In the Young's double slit experiment a monochromatic source of wavelength \(\lambda\) is used. The intensity of light passing through each slit is \(I_{0}\). The intensity of light reaching the screen \(S_{C}\) at a point \(P\), a distance \(x\) from \(O\) is given by (Take, \(d< < D\) )
- A \(I_{0} \cos ^{2}\left(\frac{\pi D}{\lambda d} x\right)\)
- B \(4 I_{0} \cos ^{2}\left(\frac{\pi d}{\lambda D} x\right)\)
- C \(I_{0} \sin ^{2}\left(\frac{\pi d}{2 \lambda D} x\right)\)
- D \(4 I_{0} \cos \left(\frac{\pi d}{2 \lambda D} x\right)\)
Answer & Solution
Correct Answer
(B) \(4 I_{0} \cos ^{2}\left(\frac{\pi d}{\lambda D} x\right)\)
Step-by-step Solution
Detailed explanation
Path difference, \(\Delta x=\frac{x d}{D}\)
So, corresponding phase difference, \(\phi=\frac{2 \pi}{\lambda}(\Delta x)\)
\(=\frac{2 \pi}{\lambda}\left(\frac{x d}{D}\right)\)
The resultant intensity at \(P\) is
\(\begin{aligned}
I_{P} &=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) \\
&=4 I_{0} \cos ^{2}\left(\frac{2 \pi}{\lambda} \frac{x d}{D} \times \frac{1}{2}\right)=4 I_{0} \cos ^{2}\left(\frac{\pi d x}{\lambda D}\right)
\end{aligned}\)
So, corresponding phase difference, \(\phi=\frac{2 \pi}{\lambda}(\Delta x)\)
\(=\frac{2 \pi}{\lambda}\left(\frac{x d}{D}\right)\)
The resultant intensity at \(P\) is
\(\begin{aligned}
I_{P} &=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) \\
&=4 I_{0} \cos ^{2}\left(\frac{2 \pi}{\lambda} \frac{x d}{D} \times \frac{1}{2}\right)=4 I_{0} \cos ^{2}\left(\frac{\pi d x}{\lambda D}\right)
\end{aligned}\)
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