Question 180: Wave Optics

Question

In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm, the angular width of central maximum is $\alpha 	imes 10^{-2}$ degrees. The value of $\alpha$ 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

In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm, the angular width of central maximum is \alpha \times 10^{-2} degrees. The value of \alpha is ________.

Integer / Numeric Answer Type