Why is the diffraction of sound waves more evident in daily experience than that of light wave?
As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm . The diffraction occur if the wavelength of waves is nearly equal to slit width. As the wavelength of light waves is $7000 \times 10^{-10} \mathrm{~m}$ to $4000 \times 10^{-10} \mathrm{~m}$. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.
The human eye has an approximate angular resolution of $\phi=5.8 \times 10^{-4}$ rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch $=2.54 \mathrm{~cm}$ ). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots.
Given, angular resolution of human eye, $\phi=5.8 \times 10^{-4} \mathrm{rad}$.
and printer prints 300 dots per inch.
The linear distance between two dots is $l=\frac{2.54}{300} \mathrm{~cm}=0.84 \times 10^{-2} \mathrm{~cm}$.
At a distance of $z \mathrm{~cm}$, this subtends an angle, $\phi=\frac{l}{z}$
$$\therefore\quad z=\frac{l}{\phi}=\frac{0.84 \times 10^{-2} \mathrm{~cm}}{5.8 \times 10^{-4}}=14.5 \mathrm{~cm}$$
A polaroid (I) is placed infront of a monochromatic source. Another polariod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.
In the diagram shown, a monochromatic light is placed infront of polaroid (I) as shown below.
As per the given question, monochromatic light emerging from polaroid (I) is plane polarised. When polaroid (II) is placed infront of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in crossed positions, i.e., pass axes of I and II are at 90$\Upsilon$
Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid II. When a third polaroid (III) is placed in between (I) and (II), no light will emerge from (II), if pass axis of (III) is parallel to pass axis of (I) or (II). In all other cases, light will emerge from (II), as pass axis of (II) will no longer be at 90 rto the pass axis of (III).
Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?
When angle of incidence is equal to Brewster's angle, the transmitted light is unpolarised and reflected light is plane polarised.
Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.
Polarisation by reflection occurs when the angle of incidence is the Brewster's angle i.e.,
$$\tan i_B={ }^1 \alpha_2=\frac{\alpha_2}{\alpha_1} \text { where } \alpha_2<\alpha_1$$
when the light rays travels in such a medium, the critical angle is
$$\sin i_c=\frac{\alpha_2}{\alpha_1}$$
where, $\alpha_2<\alpha_1$
As $\left|\tan i_B\right|>\left|\sin i_C\right|$ for large angles $i_B
Thus, the polarisation by reflection occurs definitely.
'For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000 \mathop A\limits^o$ and electrons accelerated through 100 V used as the illuminating substance.
We know that
$$\text { Resolving power }=\frac{1}{d}=\frac{2 \sin \beta}{1.22 \lambda} \Rightarrow d_{\min }=\frac{1.22 \lambda}{2 \sin \beta}$$
where, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.
For the light of wavelength $5500 \mathop A\limits^o$,
$$d_{\min }=\frac{1.22 \times 5500 \times 10^{-10}}{2 \sin \beta}\quad\text{.... (i)}$$
For electrons accelerated through 100 V , the de-Broglie wavelength
$$\begin{aligned} \lambda & =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12 \times 10^{-9} \mathrm{~m} \\ d_{\min } & =\frac{1.22 \times 0.12 \times 10^{-9}}{2 \sin \beta} \end{aligned}$$
Ratio of the least separation
$$\therefore \quad \frac{d_{\min }^{\prime}}{d_{\min }}=\frac{0.12 \times 10^{-9}}{5500 \times 10^{-10}}=0.2 \times 10^{-3}$$