Question 350: Wave Optics

Question

In a Fraunhofer diffraction at a single slit of width 'd' and incident light of wavelength $$5500 \mathop A\limits^o$$, the first minimum is observed at an angle $$30^{\circ}$$. The first secondary maxima is observed at an angle $$	heta$$, equal to

Accepted Answer

The first minimum in a Fraunhofer diffraction pattern occurs when the path difference between the light from the two edges of the slit is equal to the wavelength of the light, $$\lambda$$. The condition for the first minimum can be written as:
$$d \sin 	heta = m\lambda$$
For the first minimum, $$m = 1$$ and we know that the angle is $$30^{\circ}$$. Using the given wavelength $$\lambda = 5500 \mathop A\limits^o = 550 	imes 10^{-9}$$ meters (since $$1 \mathop A\limits^o = 10^{-10}$$ meters), we can write:
$$d \sin 30^{\circ} = 1 	imes 550 	imes 10^{-9} 	ext{ m}$$
$$d 	imes \frac{1}{2} = 550 	imes 10^{-9} 	ext{ m}$$
$$d = 2 	imes 550 	imes 10^{-9} 	ext{ m}$$
$$d = 1100 	imes 10^{-9} 	ext{ m}$$
$$d = 1100 \mathop A\limits^o$$
The secondary maxima occur in between the primary minima. The first secondary maxima (also known as the first 'bright' fringe other than the central maximum) occurs when the path difference is $$3/2$$ times the wavelength (this is the condition for the maximum that lies between the first and second minima, $$m = 1$$ and $$m = 2$$, respectively). This results in the following condition:
$$d \sin 	heta = \left(m + \frac{1}{2}ight) \lambda$$
For the first secondary maxima $$m = 1$$:
$$d \sin 	heta = \left(1 + \frac{1}{2}ight) 550 	imes 10^{-9} 	ext{ m}$$
$$d \sin 	heta = \frac{3}{2} 	imes 550 	imes 10^{-9} 	ext{ m}$$
$$1100 \sin 	heta = 3 	imes 550 	imes 10^{-9} 	ext{ m}$$
$$\sin 	heta = \frac{3}{2} 	imes \frac{550 	imes 10^{-9}}{1100 	imes 10^{-9}}$$
$$\sin 	heta = \frac{3}{2} 	imes \frac{1}{2}$$
$$\sin 	heta = \frac{3}{4}$$
Therefore, the correct answer is:
Option B
$$\sin ^{-1}\left(\frac{3}{4}ight)$$

Answer

$$\sin ^{-1}\left(\frac{1}{4}\right)$$

Answer

$$\sin ^{-1}\left(\frac{3}{4}\right)$$

Answer

$$\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$$

Answer

$$\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$$

Wave Optics

In a Fraunhofer diffraction at a single slit of width 'd' and incident light of wavelength $5500 \mathop A\limits^o, the first minimum is observed at an angle 30^{\circ}. The first secondary maxima is observed at an angle \theta$, equal to

Choose the correct answer: