Question 18: Waves Optics

Question

The wavelength of a monochromatic light which is used in single slit diffraction is 800 nm . The width of the single slit for which the first minimum appears at $	heta=45^{\circ}$ on the screen will be:

Accepted Answer

For a single slit, the first minimum occurs when
$$a\sin	heta = \lambda$$
where
Ã¢ÂÂ¢ $a$ is the slit width,
Ã¢ÂÂ¢ $\lambda=800	ext{ nm}=8.00	imes10^{-7}	ext{ m}$,
Ã¢ÂÂ¢ $	heta=45^\circ$ so $\sin45^\circ=	frac{\sqrt2}2\approx0.7071$.
Plug in the numbers:
$$a = \frac{\lambda}{\sin	heta}       = \frac{8.00	imes10^{-7}}{0.7071}      \approx1.13	imes10^{-6}	ext{ m}$$
Converting to micrometers:
$$1.13	imes10^{-6}	ext{ m} = 1.13\ \mu	ext{m}$$
That matches Option A.

Answer

$1.13 \mu \mathrm{~m}$

Answer

$1.23 \mu \mathrm{~m}$

Answer

$2.13 \mu \mathrm{~m}$

Answer

$1.3 \mu \mathrm{~m}$

Waves Optics

The wavelength of a monochromatic light which is used in single slit diffraction is 800 nm . The width of the single slit for which the first minimum appears at \theta=45^{\circ} on the screen will be:

Choose the correct answer: