Choose the correct answer from the alternatives given. The amplit… - Vidyadip

MCQ

Choose the correct answer from the alternatives given. The amplitude of an electromagnetic wave in vaccum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is correct?

A
The frequency of the wave changes only
B
The wave length of the wave changes only
C
The speed of the wave propagation changes only
✓
None of the above is correct

✓

Answer

Correct option: D.

None of the above is correct

As we know, velocity of electromagnetic wave,
$\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}=\frac{3\times10^8\text{m}}{\text{s}}$
which is constant
So It is independent of amplitude of electromagnetic wave, frequency and wavelength of electromagnetic wave.
so none of the above is correct.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Electromagnetic Waves→Electromagnetic Waves — all question types→Physics STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

A cell sends a current through a resistance $R$ for time $t$. Now the same cell sends current through another resistance $r$ for the same time. If same amount of heat is developed in both the resistance, then the internal resistance of cell is

When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor:

In a Young's Double slit experiment, first maxima is observed at a fixed point $P$ on the screen. Now the screen is continuously moved away from the plane of slits. The ratio of intensity at point $P$ to the intensity at point $O$ (centre of the screen)

When the impurity of arsenic is added to the pure semiconductor germanium, it will be present :

The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. If the thickness of the sheet is half of the separation between the plates. The capacitance will be

Two circular coils $X$ and $Y$, having equal number of turns, carry equal currents in the same sense and subtend same solid angle at point $O$. If the smaller coil, $X$ is midway between $O$ and $Y$, then if we represent the magnetic induction due to bigger coil $Y$ at $O$ as $BY$ and that due to smaller coil $X$ at $O$ as $BX$ , then

For the production of $X-$ rays, the target should be made of

Capacitance of a parallel plate capacitor becomes 4/3 times its original value if a dielectric slab of thickness t = d/2 is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is

The election field in a region is given by $\vec E = (Ax + B)\hat i$ where $E$ is in $N\,C^{-1}$ and $x$ in meters. The values of constants are $A = 20\, SI\, unit$ and $B = 10\, SI\, unit$. If the potential at $x =1$ is $V_1$ and that at $x = -5$ is $V_2$ then $V_1 -V_2$ is.....$V$