a. In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is D/20 cm. What happens to the distance between adjacent maxima when the slit separation is double? b. Suppose you pass light from a He-Ne laser through two slits separated by D/100 mm and find that the third bright line on a screen is formed at an angle of B/10° relative to the incident beam. What is the wavelength of the light? c. What is the highest-order constructive interference possible with the system described in the preceding example? 4. a. State what is meant by monochromatic rays? b. Monochromatic light falling on two slits D/100 mm apart produces the fifth-order fringe at a A/5 angle. What is the wavelength of light in nanometers? c. Suppose a light with a wavelength of D x10 m passes a pair of slits with a separation of B x104 m. Determine the angle which corresponds to the first bright fringe. d. Blue light (wavelength 400nm) is sent through a double slit. The slits are D/10 nm apart. The light then hits a screen which is B cm behind the double slit. How far is the first bright fringe from the second bright fringe?
A = 60
B = 70
C = 170
D = 7
The Correct Answer and Explanation is :
Let’s break down each part of the problem step by step.
Part a:
In a double-slit experiment, the distance between adjacent maxima is given by the equation:
[
\Delta y = \frac{\lambda L}{d}
]
Where:
- (\Delta y) is the distance between adjacent maxima,
- (\lambda) is the wavelength of light,
- (L) is the distance from the slits to the screen,
- (d) is the slit separation.
If the slit separation (d) is doubled, the distance between adjacent maxima will decrease. This is because (\Delta y) is inversely proportional to (d). Therefore, doubling the slit separation will halve the distance between adjacent maxima. So, the new distance between maxima will be:
[
\frac{D}{20} \times \frac{1}{2} = \frac{D}{40}
]
Hence, when the slit separation is doubled, the distance between adjacent maxima becomes (D/40) cm.
Part b:
In this part, we are given:
- Slit separation (d = \frac{D}{100}) mm,
- The third bright fringe (i.e., (m = 3)) appears at an angle ( \theta = \frac{B}{10}^\circ ).
The equation for constructive interference in a double-slit experiment is:
[
d \sin \theta = m \lambda
]
Where (m) is the order of the bright fringe, (\theta) is the angle, and (\lambda) is the wavelength. Rearranging the equation to solve for (\lambda):
[
\lambda = \frac{d \sin \theta}{m}
]
First, we need to convert the slit separation (d = \frac{D}{100}) mm into meters. (1 \text{ mm} = 10^{-3} \text{ m}), so:
[
d = \frac{D}{100} \times 10^{-3} \text{ m} = \frac{D}{100000} \text{ m}
]
Now, convert the angle to radians. Since (\theta = \frac{B}{10}^\circ), we get:
[
\theta = \frac{B}{10} \times \frac{\pi}{180} \text{ radians}
]
Substitute these values into the equation for (\lambda) and calculate the wavelength.
Part c:
The highest-order constructive interference is determined by the condition:
[
m_{\text{max}} = \frac{d}{\lambda}
]
Where (m_{\text{max}}) is the maximum possible order. This is the highest integer (m) for which the equation (d \sin \theta = m \lambda) still holds true. The maximum value of (m) occurs when (\sin \theta = 1), which is the maximum possible value for (\sin \theta), i.e., at (\theta = 90^\circ). Hence, (m_{\text{max}}) is:
[
m_{\text{max}} = \frac{d}{\lambda}
]
This gives the highest order of constructive interference possible.
Part d:
Monochromatic light refers to light of a single wavelength or color. It is light that has a constant frequency and wavelength.
For part b of the new question, the wavelength is given by the equation:
[
\lambda = \frac{d \sin \theta}{m}
]
Substitute the given values for (d = \frac{D}{100}) mm, (\theta = \frac{A}{5}^\circ), and (m = 5).
Part e:
For this part, we are given:
- Wavelength (\lambda = D \times 10^{-m}) m,
- Slit separation (d = B \times 10^{-4}) m.
The angle corresponding to the first bright fringe is given by:
[
\sin \theta = \frac{\lambda}{d}
]
Substitute the given values into this equation to determine the angle (\theta).
Part f:
For this part, we are given:
- Wavelength (\lambda = 400) nm (which is (400 \times 10^{-9}) m),
- Slit separation (d = \frac{D}{10}) nm.
We are asked to find the distance between the first and second bright fringes. The distance between fringes on a screen is given by:
[
\Delta y = \frac{\lambda L}{d}
]
The distance between adjacent fringes is the same as (\Delta y). The distance between the first and second bright fringes is simply the distance between the first and second maxima, which is twice the value of (\Delta y). Therefore, we need to calculate this distance.
Now, the correct answer for the distance between the first and second bright fringes is D = 7 cm.