Q: What is the phase difference for two waves of the same frequency arriving at a point with the respective path lengths of $6\lambda$ and $4.5\lambda$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$\Delta x = 6\lambda - 4.5\lambda = 1.5\lambda$, $\quad \Delta\phi = ?$
$\begin{aligned}[t] \Delta\phi &= \frac{2\pi}{\lambda} \Delta x \\ &= \frac{2\pi}{\lambda} \times 1.5\lambda = \mathbf{3\pi} \end{aligned}$
Thus, the two waves are out of phase.
Q: At what angle is the first-order maximum for $450\text{ nm}$ wavelength blue light falling on double slits separated by $0.05\text{ mm}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For double slits,
$\lambda = 450\text{ nm} = 450 \times 10^{-9}\text{ m}$,
$d = 0.05\text{ mm} = 0.05 \times 10^{-3}\text{ m}$, $\theta = ?$
For first-order bright, $m = 1$
$d\sin\theta = m\lambda$
$\sin\theta = \frac{m\lambda}{d}$
$\theta = \sin^{-1}\left(\frac{m\lambda}{d}\right)$
$\theta = \sin^{-1}\left(\frac{1 \times 450 \times 10^{-9}}{0.05 \times 10^{-3}}\right) = \mathbf{0.52^\circ}$
Q: In a double slit interference experiment, the distance between the slits is $5\text{ mm}$ and the screen is $2\text{ m}$ from the slits. Yellow light from a sodium lamp is used and it has a wavelength of $5.89 \times 10^{-7}\text{ m}$. Find the distance between the first and second order bright fringes on the screen.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For double slits,
$d = 5\text{ mm} = 5 \times 10^{-3}\text{ m}$, $D = 2\text{ m}$
$\lambda = 5.89 \times 10^{-7}\text{ m}$, $\Delta y = ?$
$\begin{aligned}[t] \Delta y &= \frac{\lambda D}{d} \\ &= \frac{5.89 \times 10^{-7} \times 2}{5 \times 10^{-3}} \\ &= 2.356 \times 10^{-4}\text{ m} = \mathbf{0.2356\text{ mm}} \end{aligned}$
Q: Find the distance between adjacent dark spots from a double slit interference pattern if the wavelength of light is $500\text{ nm}$, the distance between the slits is $1\text{ mm}$, and the distance from the slit to the screen is $2\text{ m}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For double slits,
$d = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$, $D = 2\text{ m}$,
$\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}$, $\Delta y = ?$
$\begin{aligned}[t] \Delta y &= \frac{\lambda D}{d} \\ &= \frac{500 \times 10^{-9} \times 2}{1 \times 10^{-3}} \\ &= 0.001\text{ m} = \mathbf{1\text{ mm}} \end{aligned}$
Q: If the distance between two slits is $0.05\text{ mm}$ and the distance to a screen is $2.5\text{ m}$, find the spacing between the first-order and second-order bright fringes for yellow light of $580\text{ nm}$ wavelength.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For double slits,
$d = 0.05\text{ mm} = 0.05 \times 10^{-3}\text{ m}$, $D = 2.5\text{ m}$,
$\lambda = 580\text{ nm} = 580 \times 10^{-9}\text{ m}$, $\Delta y = ?$
$\begin{aligned}[t] \Delta y &= \frac{\lambda D}{d} \\ &= \frac{580 \times 10^{-9} \times 2.5}{0.05 \times 10^{-3}} \\ &= 0.029\text{ m}\text{ (or) } \mathbf{29\text{ mm}} \end{aligned}$
Q: A beam of $650\text{ nm}$ light is directed upon two slits that are separated by a distance of $0.75\text{ mm}$. A screen is placed $1.50\text{ m}$ away to capture the interference pattern. What is the distance from the $1^{\text{st}}$ order maximum to the $3^{\text{rd}}$ order minimum?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For double slits,
$d = 0.75\text{ mm} = 0.75 \times 10^{-3}\text{ m}$, $D = 1.5\text{ m}$,
$\lambda = 650\text{ nm} = 650 \times 10^{-9}\text{ m}$, $\Delta y = ?$
For $1^{\text{st}}$ order bright (maximum), $m = 1$
$\begin{aligned}[t]
y_1 &= m \frac{\lambda D}{d} \\
&= \frac{650 \times 10^{-9} \times 1.5}{0.75 \times 10^{-3}} \\
&= 0.0013\text{ m} = 1.3 \times 10^{-3}\text{ m}
\end{aligned}$
For $3^{\text{rd}}$ order dark (minimum), $m = 2$
$\begin{aligned}[t]
y_2 &= \left(m + \frac{1}{2}\right) \frac{\lambda D}{d} = \left(2 + \frac{1}{2}\right) \frac{\lambda D}{d} = \frac{5}{2} \frac{\lambda D}{d} \\
&= \frac{5}{2} \times \frac{650 \times 10^{-9} \times 1.5}{0.75 \times 10^{-3}} \\
&= 3.25 \times 10^{-3}\text{ m}
\end{aligned}$
$\begin{aligned}[t] \Delta y &= y_2 - y_1 \\ &= 3.25 \times 10^{-3} - 1.3 \times 10^{-3} \\ &= \mathbf{1.95 \times 10^{-3}\text{ m}} \end{aligned}$
Q: A beam of green light from a laser light source is diffracted by a slit of width $0.55\text{ mm}$. The diffraction pattern forms on a wall $2.06\text{ m}$ beyond the slit. The distance between the positions of the first minima on both sides of central bright fringe is $4.1\text{ mm}$. Calculate the wavelength of the laser light.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For single slit,
$a = 0.55\text{ mm} = 0.55 \times 10^{-3}\text{ m}$,
$D = 2.06\text{ m}$, $\lambda = ?$
The width of the central maximum $= 2\frac{\lambda D}{a}$
$4.1\text{ mm} = 4.1 \times 10^{-3}\text{ m}$
$\begin{aligned}[t]
4.1 \times 10^{-3} &= 2 \frac{\lambda D}{a} \\
\lambda &= \frac{4.1 \times 10^{-3} \times a}{2D} \\
&= \frac{4.1 \times 10^{-3} \times 0.55 \times 10^{-3}}{2 \times 2.06} \\
&= 547.3 \times 10^{-9}\text{ m} = \mathbf{547.3\text{ nm}}
\end{aligned}$
Q: Light of wavelength $580\text{ nm}$ is incident on a slit of width $0.3\text{ mm}$. The observing screen is placed $2\text{ m}$ from the slit. Find the positions of the first dark fringes and the width of the central bright fringes.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For single slit,
$a = 0.3\text{ mm} = 0.3 \times 10^{-3}\text{ m}$, $D = 2\text{ m}$,
$\lambda = 580\text{ nm} = 580 \times 10^{-9}\text{ m}$, $y = ?$
For first dark fringes, $m = 1$
$\begin{aligned}[t]
y &= m \frac{\lambda D}{a} \\
&= \frac{580 \times 10^{-9} \times 2}{0.3 \times 10^{-3}} \\
&= \mathbf{3.87 \times 10^{-3}\text{ m}}
\end{aligned}$
The width of the central maximum $= 2\frac{\lambda D}{a}$
$\begin{aligned}[t]
&= \frac{2 \times 580 \times 10^{-9} \times 2}{0.3 \times 10^{-3}} \\
&= 7.73 \times 10^{-3}\text{ m} = \mathbf{7.74\text{ mm}}
\end{aligned}$
[OR] The width of the central maximum
$= 2y = 2 \times 3.87 \times 10^{-3} = \mathbf{7.74\text{ mm}}$
Q: Light of wavelength $457\text{ nm}$ illuminates a diffraction grating with $5000\text{ lines cm}^{-1}$. What is the second order diffracted angle?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$\lambda = 457\text{ nm} = 457 \times 10^{-9}\text{ m}$,
$N = 5000\text{ lines cm}^{-1}$, $\theta = ?$
for second order, $m = 2$
$\begin{aligned}[t] d &= \frac{1}{N} \\ &= \frac{1}{5000} = 2 \times 10^{-4}\text{ cm} \\ &= 2 \times 10^{-6}\text{ m} \end{aligned}$
$d\sin\theta = m\lambda$
$\sin\theta = \frac{m\lambda}{d}$
$\theta = \sin^{-1}\left(\frac{m\lambda}{d}\right)$
$\theta = \sin^{-1}\left(\frac{2 \times 457 \times 10^{-9}}{2 \times 10^{-6}}\right) = \mathbf{27.19^\circ\text{ (or) } 27^\circ\text{ 11'}}$
Q: A diffraction grating has $2000\text{ lines per centimeter}$. At what angle will the first-order maximum be for $520\text{ nm}$ wavelength green light?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$\lambda = 520\text{ nm} = 520 \times 10^{-9}\text{ m}$,
$N = 2000\text{ lines cm}^{-1}$, $\theta = ?$
for first-order, $m = 1$
$\begin{aligned}[t] d &= \frac{1}{N} \\ &= \frac{1}{2000} = 5 \times 10^{-4}\text{ cm} \\ &= 5 \times 10^{-6}\text{ m} \end{aligned}$
$d\sin\theta = m\lambda$
$\sin\theta = \frac{m\lambda}{d}$
$\theta = \sin^{-1}\left(\frac{m\lambda}{d}\right)$
$\theta = \sin^{-1}\left(\frac{1 \times 520 \times 10^{-9}}{5 \times 10^{-6}}\right) = \mathbf{5.97^\circ\text{ (or) } 5^\circ\text{ 58'}}$
Q: A grating containing $4000\text{ slits per centimeter}$ is illuminated with monochromatic light and produces the second-order bright line at a $30^\circ$ angle. What is the wavelength of the light used?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$N = 4000\text{ lines cm}^{-1}$, $\theta = 30^\circ$, $\lambda = ?$
for second order, $m = 2$
$\begin{aligned}[t] d &= \frac{1}{N} \\ &= \frac{1}{4000} = 2.5 \times 10^{-4}\text{ cm} \\ &= \mathbf{2.5 \times 10^{-6}\text{ m}} \end{aligned}$
$d\sin\theta = m\lambda$
$\begin{aligned}[t]
\lambda &= \frac{d\sin\theta}{m} \\
&= \frac{2.5 \times 10^{-6} \times \sin 30^\circ}{2} \\
&= \mathbf{625 \times 10^{-9}\text{ m} = 625\text{ nm}}
\end{aligned}$
Q: How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for $470\text{ nm}$ blue light at an angle of $25^\circ$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$N = ?$, $\lambda = 470\text{ nm} = 470 \times 10^{-9}\text{ m}$,
$\theta = 25^\circ$
for first-order, $m = 1$
$d\sin\theta = m\lambda$
$\begin{aligned}[t]
d &= \frac{m\lambda}{\sin\theta} \\
&= \frac{1 \times 470 \times 10^{-9}}{\sin 25^\circ} \\
&= 1.112 \times 10^{-6}\text{ m} \\
&= \mathbf{1.112 \times 10^{-4}\text{ cm}}
\end{aligned}$
$\begin{aligned}[t] N &= \frac{1}{d} \\ &= \frac{1}{1.112 \times 10^{-4}} \\ &= \mathbf{8992\text{ lines cm}^{-1}} \end{aligned}$