Electromagnetic Waves MCQ Questions & Answers in Electrostatics and Magnetism | Physics

It is an electromagnetic wave. The frequency of the microwave oven must match the resonant frequency of the water molecules.
NOTE
In the presence of microwaves, water molecules oscillate with frequency of microwaves and large heat is developed.

Wave is uv rays.

$$\eqalign{ & E = \frac{{hc}}{\lambda } \Rightarrow \lambda = \frac{{hc}}{E} \cr & \Rightarrow \lambda = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{11 \times 1000 \times 1.6 \times {{10}^{ - 19}}}} = 12.4\,\mathop {\text{A}}\limits^ \circ \cr} $$

wavelength range of visible region is $$4000\,\mathop {\text{A}}\limits^ \circ $$  to $$7800\,\mathop {\text{A}}\limits^ \circ .$$

If maximum electron density of the ionosphere is $${N_{\max }}$$ per $${m^3}$$ then the critical frequency $${f_c}$$ is given by $${f_c} = 9{\left( {{N_{\max }}} \right)^{\frac{1}{2}}}$$
$$\eqalign{ & \Rightarrow 10 \times {10^6} = 9{\left( N \right)^{\frac{1}{2}}} \cr & \Rightarrow N = 1.2 \times {10^{12}}{m^{ - 3}} \cr} $$

Infrared causes heating effect.

$$\eqalign{ & \upsilon = \frac{1}{{2\pi \sqrt {LC} }} \cr & \lambda = \frac{C}{\upsilon } \Rightarrow \lambda = C \times 2\pi \sqrt {LC} \cr & \lambda = 3 \times {10^8} \times 2 \times 3.14 \times \sqrt {10 \times {{10}^{ - 3}} \times 4 \times {{10}^{ - 12}}} \cr & = 377\,m \cr} $$

Intersity of a plane electromagnetic wave
$$ = \frac{1}{2}{\varepsilon _0}E_0^2c,\,{\text{so}}\,E = \sqrt {\frac{{2I}}{{{\varepsilon _0}c}}} = 38.8\,N{C^{ - 1}}$$

Ozone layer extends from $$30\,km$$  to nearly $$50\,km$$  above the earth's surface in ozone sphere. This layer absorbs the major part of ultraviolet tadiations coming from the sun and does not allow them to reach the earth's surface.
The range of ultraviolet radiations is $$100\,\mathop {\text{A}}\limits^ \circ $$  to $$4000\,\mathop {\text{A}}\limits^ \circ .$$  Thus, it blocks the radiations of wavelength less than $$3 \times {10^{ - 7}}m\left( {{\text{or}}\,\,3000\,\mathop {\text{A}}\limits^ \circ } \right).$$

In vacuum velocity of all $$EM$$  waves are same but their wavelengths are different.

Frequency remains constant during refraction
$$\eqalign{ & {v_{{\text{med}}}} = \frac{1}{{\sqrt {{\mu _0}{ \in _0} \times 4} }} = \frac{c}{2} \cr & \frac{{{\lambda _{{\text{med}}}}}}{{{\lambda _{{\text{air}}}}}} = \frac{{{v_{{\text{med}}}}}}{{{v_{{\text{air}}}}}} = \frac{{\frac{c}{2}}}{c} = \frac{1}{2} \cr} $$
$$\therefore $$ wavelength is halved and frequency remains unchanged.