Question 94: Wave Optics

Question

The ratio of intensities at two points $$\mathrm{P}$$ and $$\mathrm{Q}$$ on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are $$\pi / 3$$ and $$\pi / 2$$, respectively are

Accepted Answer

In a Young's double slit experiment, the intensity at a point on the screen is given by the formula :
$$I = 4I_0\cos^2\left(\frac{\pi d\sin	heta}{\lambda}ight)$$
where $I_0$ is the intensity at the center of the pattern, $d$ is the distance between the slits, $	heta$ is the angle between the line joining the point to the center of the pattern and the line passing through the center of the pattern and the slits, and $\lambda$ is the wavelength of the light used.
The phase difference between the waves from the two slits at a point on the screen is given by :
$$\Delta \phi = \frac{2\pi d\sin	heta}{\lambda}$$
For a phase difference of $\pi/3$ between the waves from the two slits at point P, we have :
$$\frac{\Delta \phi}{\pi} = \frac{2d\sin	heta}{\lambda} = \frac{1}{3}$$
For a phase difference of $\pi/2$ between the waves from the two slits at point Q, we have :
$$\frac{\Delta \phi}{\pi} = \frac{2d\sin	heta}{\lambda} = \frac{1}{2}$$
Solving for $\sin	heta$ in both cases, we get:
$$\sin	heta_P = \frac{\lambda}{6d},\quad \sin	heta_Q = \frac{\lambda}{4d}$$
Substituting these values in the formula for intensity, we get :
$$\frac{I_P}{I_Q} = \frac{4\cos^2\left(\frac{\pi}{6}ight)}{4\cos^2\left(\frac{\pi}{4}ight)} = \frac{3}{2}$$
Therefore, the ratio of intensities at points P and Q is 3 : 2

Answer

2 : 3

Answer

1 : 3

Answer

3 : 1

Answer

3 : 2

Wave Optics

The ratio of intensities at two points $\mathrm{P} and \mathrm{Q} on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are \pi / 3 and \pi / 2$, respectively are

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