Question bank/JEE Main/Physics

JEE MainPhysicsModerateWave Optics

In a double-slit experiment, at a certain point on the screen
the path difference between the two interfering waves is

18\frac{1}{8}thof a wavelength. The ratio of the intensity of light at
that point to that at the centre of a bright fringe is

Show answer and solution
I=I0cos⁡2(Δϕ2)I = I_{0}\cos^{2}(\frac{\Delta\phi}{2})
II0=cos⁡2[2πλ×Δx2]=cos⁡2(π8)\frac{I}{I_{0}} = \cos^{2}\lbrack\frac{\frac{2\pi}{\lambda} \times \Delta x}{2}\rbrack = \cos^{2}(\frac{\pi}{8});
II0=0.853\frac{I}{I_{0}} = 0.853

Source: Grass-G / P-19 Physics paper 28 Nov Fourth.docx

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