Question 179: Wave Optics

Question

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is _______.

Accepted Answer

For a single slit diffraction pattern, the angular width of the central maximum is
$$ 2	heta $$
where the first minima occur at
$$ a \sin	heta = \lambda $$
For very small angles, we use
$$ \sin	heta \approx 	heta $$
So,
$$ 	heta \approx \frac{\lambda}{a} $$
Hence angular width of central maximum is
$$ 2	heta = \frac{2\lambda}{a} $$
Now substitute the given values:
$$ \lambda = 628\ 	ext{nm} = 628 	imes 10^{-9}\ 	ext{m} $$
$$ a = 0.2\ 	ext{mm} = 0.2 	imes 10^{-3}\ 	ext{m} = 2 	imes 10^{-4}\ 	ext{m} $$
Therefore,
$$ 2	heta = \frac{2 	imes 628 	imes 10^{-9}}{2 	imes 10^{-4}} $$
$$ 2	heta = 628 	imes 10^{-5}\ 	ext{rad} $$
$$ 2	heta = 6.28 	imes 10^{-3}\ 	ext{rad} $$
Now convert into degrees:
$$ 1\ 	ext{rad} = \frac{180}{\pi}^\circ $$
So,
$$ 2	heta = 6.28 	imes 10^{-3} 	imes \frac{180}{\pi} $$
$$ 2	heta \approx 0.36^\circ $$
Given that angular width is $\alpha 	imes 10^{-2}$ degrees,
$$ 0.36^\circ = 36 	imes 10^{-2}\ ^\circ $$
So,
$$ \alpha = 36 $$
The answer is:
$$ \boxed{36} $$

Wave Optics

Integer / Numeric Answer Type